Derivative as a function - AP Calculus AB

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Question

A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after seconds. What's the acceleration in $m/s^{2}$ of the block after it has been ejected?

Answer

Since $a=\frac{ds^2}{dt^2}$ , by differentiating the position function twice, we see that acceleration is constant and $-32\ m/s^{2}$ . Acceleration, in this case, is gravity, which makes sense that it should be a constant value!

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Question

A jogger leaves City at . His subsequent position, in feet, is given by the function:

$x(t) = 5t^{2} - 3t + 2$ ,

where is the time in minutes.

Find the acceleration of the jogger at minutes.

Answer

The accelaration is given by the second derivative of the position function:

For the given position function:

$f(t) = 5t^{2} - 3t + 2$ ,

,

.

Therefore, the acceleration at minutes is $10\ ft/min^{2}$ . Again, note the units must be in $ft/min^{2}$ .

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Question

The speed of a car traveling on the highway is given by the following function of time:

Consider a second function:

What can we conclude about this second function?

Answer

Notice that the function is simply the derivative of with respect to time. To see this, simply use the power rule on each of the two terms.

Therefore, is the rate at which the car's speed changes, a quantity called acceleration.

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Question

Find the critical numbers of the function,

$f(x)=\frac{1+x^2}{x}$

Answer

$f(x)=\frac{1+x^2}{x}$

1) Recall the definition of a critical point:

The critical points of a function are defined as points , such that is in the domain of , and at which the derivative is either zero or does not exist. The number is called a critical number of .

2) Differentiate ,

$f'(x)=\frac{d}{dx}\left(\frac{1+x^2}{x} \right )=\frac{d}{dx}\left(\frac{1}{x}+x \right )=-\frac{1}{x^2}+1$

3) Set to zero and solve for ,

$f'(x)=-\frac{1}{x^2}+1= 0$

$\frac{1}{x^2}=1$

The critical numbers are,

$x =\left { -1,1\right }$

We can also observe that the derivative does not exist at , since the $-\frac{1}{x^2}$ term would be come infinite. However, is not a critical number because the original function is not defined at . The original function is infinite at , and therefore is a vertical asymptote of as can be seen in its' graph,

Further Discussion

In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function at the critical numbers to find the corresponding function values and then write them as a set of ordered pairs,

$\left {(-1,-2), (1,2) \right }$

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Question

The function is a continuous, twice-differentiable functuon defined for all real numbers.

If the following are true:

Which function could be ?

Answer

To answer this problem we must first interpret our given conditions:

Implies the function is strictly increasing.

Implies the function is strictly concave down.

We note the only function given which fufills both of these conditions is .

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Question

A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after seconds. What's the acceleration in $m/s^{2}$ of the block after it has been ejected?

Answer

Since $a=\frac{ds^2}{dt^2}$ , by differentiating the position function twice, we see that acceleration is constant and $-32\ m/s^{2}$ . Acceleration, in this case, is gravity, which makes sense that it should be a constant value!

Compare your answer with the correct one above

Question

A jogger leaves City at . His subsequent position, in feet, is given by the function:

$x(t) = 5t^{2} - 3t + 2$ ,

where is the time in minutes.

Find the acceleration of the jogger at minutes.

Answer

The accelaration is given by the second derivative of the position function:

For the given position function:

$f(t) = 5t^{2} - 3t + 2$ ,

,

.

Therefore, the acceleration at minutes is $10\ ft/min^{2}$ . Again, note the units must be in $ft/min^{2}$ .

Compare your answer with the correct one above

Question

The speed of a car traveling on the highway is given by the following function of time:

Consider a second function:

What can we conclude about this second function?

Answer

Notice that the function is simply the derivative of with respect to time. To see this, simply use the power rule on each of the two terms.

Therefore, is the rate at which the car's speed changes, a quantity called acceleration.

Compare your answer with the correct one above

Question

Find the critical numbers of the function,

$f(x)=\frac{1+x^2}{x}$

Answer

$f(x)=\frac{1+x^2}{x}$

1) Recall the definition of a critical point:

The critical points of a function are defined as points , such that is in the domain of , and at which the derivative is either zero or does not exist. The number is called a critical number of .

2) Differentiate ,

$f'(x)=\frac{d}{dx}\left(\frac{1+x^2}{x} \right )=\frac{d}{dx}\left(\frac{1}{x}+x \right )=-\frac{1}{x^2}+1$

3) Set to zero and solve for ,

$f'(x)=-\frac{1}{x^2}+1= 0$

$\frac{1}{x^2}=1$

The critical numbers are,

$x =\left { -1,1\right }$

We can also observe that the derivative does not exist at , since the $-\frac{1}{x^2}$ term would be come infinite. However, is not a critical number because the original function is not defined at . The original function is infinite at , and therefore is a vertical asymptote of as can be seen in its' graph,

Further Discussion

In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function at the critical numbers to find the corresponding function values and then write them as a set of ordered pairs,

$\left {(-1,-2), (1,2) \right }$

Compare your answer with the correct one above

Question

The function is a continuous, twice-differentiable functuon defined for all real numbers.

If the following are true:

Which function could be ?

Answer

To answer this problem we must first interpret our given conditions:

Implies the function is strictly increasing.

Implies the function is strictly concave down.

We note the only function given which fufills both of these conditions is .

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Question

When , what is the concavity of the graph of $2x^4+\frac{1}{2}x$ ?

Answer

To find the concavity, we need to look at the first and second derivatives at the given point.

To take the first derivative of this equation, use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{4-1})+(1\frac{1}{2}x^{1-1})$

Simplify:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{3})+(\frac{1}{2}x^{0})$

Remember that anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(8x^{3})+(\frac{1}{2}1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=8x^{3}+\frac{1}{2}$

The first derivative tells us if the function is increasing or decreasing. Plug in the given point, , to see if the result is positive (i.e. increasing) or negative (i.e. decreasing).

$y=8(3)^{3}+\frac{1}{2}$

Therefore the function is increasing.

To find out if the function is convex, we need to look at the second derivative evaluated at the same point, , and check if it is positive or negative.

We're going to treat $\frac{1}{2}$ as $\frac{1}{2}x^0$ since anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(38x^{3-1})+(0\frac{1}{2}x^{0-1})$

Notice that $(0\frac{1}{2}x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(38x^{2})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=24x^{2}$

Plug in our given value:

Since the second derivative is positive, the function is convex.

Therefore, we are looking at a graph that is both increasing and convex at our given point.

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Question

At the point , is increasing or decreasing, and is it concave or convex?

Answer

To find out if the function is increasing or decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{2-1})+(112x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{1})+(112x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

Now we plug in our given value and find out if the result is positive or negative. If it is positive, the function is increasing. If it is negative, the function is decreasing.

Therefore, the function is decreasing.

To find out if it is concave or convex, look at the second derivative. If the result is positive, it is convex. If it is negative, then it is concave.

To find the second derivative, we repeat the process using as our expression.

We're going to treat as .

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(110x^{1-1})+(012x^{0-1})$

Notice that $(012x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(110x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(10x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=10$

Since we get a positive constant, it doesn't matter where we look on the graph, as our second derivative will always be positive. That means that this graph is going to be convex at our given point.

Therefore, the function is decreasing and convex at our given point.

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Question

At the point , is the function increasing or decreasing, concave or convex?

Answer

First, let's find out if the graph is increasing or decreasing. For that, we need the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2-8x^{2-1})+(015x^{0-1})$

Notice that $(015x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

Plug in our given point for . If the result is positive, the function is increasing. If the result is negative, the function is decreasing.

Our result is negative, therefore the function is decreasing.

To find the concavity, look at the second derivative. If the function is positive at our given point, it is concave. If the function is negative, it is convex.

To find the second derivative we repeat the process, but using as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=1*-16x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16x^{0}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16$

As you can see, our second derivative is a constant. It doesn't matter what point we plug in for ; our output will always be negative. Therefore our graph will always be convex.

Combine our two pieces of information to see that at the given point, the graph is decreasing and convex.

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Question

When , what is the concavity of the graph of $2x^4+\frac{1}{2}x$ ?

Answer

To find the concavity, we need to look at the first and second derivatives at the given point.

To take the first derivative of this equation, use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{4-1})+(1\frac{1}{2}x^{1-1})$

Simplify:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{3})+(\frac{1}{2}x^{0})$

Remember that anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(8x^{3})+(\frac{1}{2}1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=8x^{3}+\frac{1}{2}$

The first derivative tells us if the function is increasing or decreasing. Plug in the given point, , to see if the result is positive (i.e. increasing) or negative (i.e. decreasing).

$y=8(3)^{3}+\frac{1}{2}$

Therefore the function is increasing.

To find out if the function is convex, we need to look at the second derivative evaluated at the same point, , and check if it is positive or negative.

We're going to treat $\frac{1}{2}$ as $\frac{1}{2}x^0$ since anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(38x^{3-1})+(0\frac{1}{2}x^{0-1})$

Notice that $(0\frac{1}{2}x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(38x^{2})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=24x^{2}$

Plug in our given value:

Since the second derivative is positive, the function is convex.

Therefore, we are looking at a graph that is both increasing and convex at our given point.

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Question

At the point , is increasing or decreasing, and is it concave or convex?

Answer

To find out if the function is increasing or decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{2-1})+(112x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{1})+(112x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

Now we plug in our given value and find out if the result is positive or negative. If it is positive, the function is increasing. If it is negative, the function is decreasing.

Therefore, the function is decreasing.

To find out if it is concave or convex, look at the second derivative. If the result is positive, it is convex. If it is negative, then it is concave.

To find the second derivative, we repeat the process using as our expression.

We're going to treat as .

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(110x^{1-1})+(012x^{0-1})$

Notice that $(012x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(110x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(10x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=10$

Since we get a positive constant, it doesn't matter where we look on the graph, as our second derivative will always be positive. That means that this graph is going to be convex at our given point.

Therefore, the function is decreasing and convex at our given point.

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Question

At the point , is the function increasing or decreasing, concave or convex?

Answer

First, let's find out if the graph is increasing or decreasing. For that, we need the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2-8x^{2-1})+(015x^{0-1})$

Notice that $(015x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

Plug in our given point for . If the result is positive, the function is increasing. If the result is negative, the function is decreasing.

Our result is negative, therefore the function is decreasing.

To find the concavity, look at the second derivative. If the function is positive at our given point, it is concave. If the function is negative, it is convex.

To find the second derivative we repeat the process, but using as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=1*-16x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16x^{0}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16$

As you can see, our second derivative is a constant. It doesn't matter what point we plug in for ; our output will always be negative. Therefore our graph will always be convex.

Combine our two pieces of information to see that at the given point, the graph is decreasing and convex.

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Question

Practicing the chain rule level 1 E!

Find the derivative of the function

Answer

To understand why the answer is

$y(x)'=\frac{4}{4x-3}$ ,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

And finally, you must understand that the derivative of

is actually

.

$y(x)=ln(\sqrt{sin(x^2)})$ can be treated as a composition of the functions

, $g(x)=\sqrt{sin(x)}$ and .

The reason why we did not define $g(x)=\sqrt{x}$ , and

is because of the property mentioned before,

turning

$ln(\sqrt{g(h(x))})$

into

$\frac{1}{2}ln(g(h(x)))$

in terms of , and is actually which means

$f(g(h(x)))=\frac{1}{2}ln(sin(x^2))$ since in is substituted with and in can be substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Look at the next function h(x), keep it inside the other function g(x)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))g'(h(x))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\frac{1}{2}\frac{1}{sin(x^2)}cos(x^2)*2x$ which is

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

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Question

Practicing the chain rule level 3 A!

Find the derivative of the function

$y(x)=ln(\sqrt{sin(x^2)})$

Answer

To understand why the answer is

,

first remember that .

Then, you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

And finally, you must understand that the derivative of

is actually

.

$y(x)=ln(\sqrt{sin(x^2)})$ can be treated as a composition of the functions

, $g(x)=\sqrt{sin(x)}$ and .

The reason why we did not define $g(x)=\sqrt{x}$ , and

is because of the property mentioned before,

turning

$ln(\sqrt{g(h(x))})$

into

$\frac{1}{2}ln(g(h(x)))$

in terms of , and is actually which means

$f(g(h(x)))=\frac{1}{2}ln(sin(x^2))$ since in is substituted with and in can be substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Look at the next function h(x), keep it inside the other function g(x)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))g'(h(x))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\frac{1}{2}\frac{1}{sin(x^2)}cos(x^2)*2x$ which is

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

Compare your answer with the correct one above

Question

Practicing the chain rule level 3 B!

Find the derivative of the function

Answer

To understand why the answer is

$q(z)=cos(cos(tan(z)))sin(tan(z))sec^{2}(z)$ ,

you must understand that the derivative of

is actually

.

Next, you must understand that the derivative of

is actually

.

And finally, you must understand that the derivative of

is actually

$q'(z)=sec^{2}(z)$ .

can be treated as a composition of the functions

, and .

in terms of , and is actually which means

since in is substituted with and in can be substituted with .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(z) first, then differentiate it

Step 2: Look at the next function g(z), keep it inside the other function f'(z)

Step 3: Look at the next function h(z), keep it inside the other function g(z)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(z) but multiply it by the factors f'(g(h(z)))g'(h(z))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(z), g(z), g'(z), h(z) and h'(z) for the expressions you found before:

$cos(cos(tan(z)))sin(tan(z)))*sec^{2}(z)$

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to differentiate. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

Compare your answer with the correct one above

Question

Practicing the chain rule level 3 C!

Find the derivative of the function

$y(x)=\frac{1}{ln((x^2+3)^{-2})}$

Answer

To understand why the answer is

$y'(x)=\frac{4x}{(x^2+3)ln^2(\frac{1}{(x^2+3)^2})}$ ,

Then, you must understand that the derivative of

$\large y(x)=\frac{1}{x}$

is actually

$\large y'(x)=\frac{-1}{x^2}$ .

Next, you must understand that the derivative of

$\large y(x)=ln(x)$

is actually

$\large y'(x)=\frac{1}{x}$ .

And finally, you must understand that the derivative of

$\frac{1}{(2x+3)^2}$

is actually

$y'(x)=\frac{-4x}{(x^2+3)^3}$ .

$y(x)=\frac{1}{ln((x^2+3)^{-2})}$ can be treated as a composition of the functions

$f(x)=\frac{1}{x+a}$ , and .

in terms of , and is actually which means

$f(g(h(x)))=\frac{1}{ln((x^2+3)^{-2})}$ since in $\frac{1}{x+a}$ is substituted with and in can be substituted with $\frac{1}{(x^2+3)^2}$ .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

Step 3: Look at the next function h(x), keep it inside the other function g(x)

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))g'(h(x))

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\large \large y'(x)=\frac{4x}{(x^2+3)*ln^2(\frac{1}{(x^2+3)^2})}$

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

then it's derivative would be

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

Compare your answer with the correct one above

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