Calculus I — Derivatives - High School Math

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Question

Calculate $\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

Answer

You can substitute $u = 4x^{2}$ to write this as:

$\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

$= \lim_{x \rightarrow 0}\left (4 \cdot \frac{\sin 4 x^{2} }{4x^{2}} \right )$

$=4 \lim_{x \rightarrow 0} \frac{\sin 4 x^{2} }{4x^{2}}$

$=4 \lim_{u \rightarrow 0} \frac{\sin u }{u}$

Note that as $x \rightarrow 0$ , $u = 4x^{2} \rightarrow 0$

$\lim_{u \rightarrow 0} \frac{\sin u }{u}$ , since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

$\lim_{u \rightarrow 0} \frac{\sin u }{u}=\frac{(\sin u)'}{u'}=\frac{\cos u}{1}=\cos u$

$4\lim_{u\rightarrow 0}\cos u=4\cdot \cos(0)=4\cdot 1$

$4 \lim_{u \rightarrow 0} \frac{\sin u }{u} = 4 \cdot 1 = 4$ , which is the correct choice.

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Question

Calculate $\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$ .

Answer

Substitute $u = 2x^{2} - 2$ to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as $x \rightarrow1$ , $u \rightarrow 0$ .)

$\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$

$=2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2\left (x^{2}-1 \right )}$

$= 2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2x^{2}-2 \right )}$

$= 2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u}$

$\lim_{u \rightarrow 0} \frac{\sin u }{u} = 1$ , so

$2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u} = 2\cdot1 = 2$ , which is therefore the correct answer choice.

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Question

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

Note that

What does this mean?

Answer

The function gives you the car's speed at time . Therefore, the fact that means that the car's speed is at time . This is equivalent to saying that the car is not moving at time . We have to take the derivative of to make claims about the acceleration.

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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 $f^{'}(2)$ if the function is given by

Answer

To find the derivative at , we first take the derivative of . By the derivative rule for logarithms,

$f'(x) = \frac{1}{x}$

Plugging in , we get

$f'(2) = \frac{1}{2}$

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Question

Find the derivative of the following function at the point .

Answer

Use the power rule on each term of the polynomial to get the derivative,

Now we plug in

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Question

Let $f(x)=x^2\sin(x^2)$ . What is $f'\left (\frac{\sqrt{\pi}}{2} \right )$ ?

Answer

We need to find the first derivative of f(x). This will require us to apply both the Product and Chain Rules. When we apply the Product Rule, we obtain:

$f'(x)=\sin(x^2)\cdot\frac{\mathrm{d} }{\mathrm{d} x}[x^2]+x^2\cdot\frac{\mathrm{d} }{\mathrm{d} x}[\sin(x^2)]$

In order to find the derivative of $\sin(x^2)$ , we will need to employ the Chain Rule.

$\frac{\mathrm{d} }{\mathrm{d} x}[\sin(x^2)]=\cos(x^2)\cdot\frac{\mathrm{d} }{\mathrm{d} x}[x^2]=\cos(x^2)\cdot2x$

$f'(x)=\sin(x^2)\cdot2x + x^2\cdot\cos(x^2)\cdot2x$

We can factor out a 2x to make this a little nicer to look at.

$f'(x)=2x(\sin(x^2)+x^2\cos(x^2))$

Now we must evaluate the derivative when x = $\frac{\sqrt{\pi}}{2}$ .

$f'\left (\frac{\sqrt{\pi}}{2} \right )=2\cdot\frac{\sqrt{\pi}}{2}(\sin\frac{\pi}{4}+\frac{\pi}{4}\cos\frac{\pi}{4})$

$=\sqrt{\pi}(\frac{\sqrt{2}}{2}+\frac{\pi\sqrt{2}}{8})=\sqrt{2\pi}(\frac{1}{2}+\frac{\pi}{8})=\frac{\sqrt{2\pi}(\pi + 4)}{8}$

The answer is $\frac{\sqrt{2\pi}(\pi + 4)}{8}$ .

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Question

Find the derivative of the following function:

Answer

We use the power rule on each term of the function.

The first term

becomes

$2 \times 6 x^2 = 12 x$ .

The second term

becomes

.

The final term, 7, is a constant, so its derivative is simply zero.

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Question

Give the average rate of change of the function $f(x) = 4 ^{x}$ on the interval .

Answer

The average rate of change of on interval is

$\frac{f (b ) - f(a)}{b-a}$

Substitute:

$\frac{f (4 ) - f(3)}{4-3} = \frac{4^{4} - 4^{3}}{1} = \frac{256 - 64}{1} = 192$

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Question

What is the derivative of ?

Answer

To get $\frac{\mathrm{d} }{\mathrm{d} x}(2x)$ , we can use the power rule.

Since the exponent of the is , as , we lower the exponent by one and then multiply the coefficient by that original exponent:

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

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=2x^{0}$

Anything to the power is .

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=21$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=2$

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Question

What is the derivative of ?

Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2x^{2-1})+(15x^{1-0})$

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

Remember that anything to the zero power is one.

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

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

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Question

What is the derivative of ?

Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

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

That means this problem will look like this:

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})+(08x^{0-1})$

Notice that $(08x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{0})$

Remember, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=51$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5$

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Question

What is the derivative of ?

Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

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

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})+(05x^{0-1})$

Notice that $(05x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{2})+(2x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=3x^2+2$

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Question

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=?$

Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

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

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})+(08x^{0-1})$

Notice that $08x^{0-1}=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=12x$

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Question

What is the derivative of ?

Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

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

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

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

That leaves us with $\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})$ .

Simplify.

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

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(x^{0})$

As stated earlier, anything to the zero power is one, leaving us with:

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=1$

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Question

What is the derivative of ?

Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

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

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})+(04x^{0-1})$

Notice that $(04x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{1})+(113x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x^{1})+(13x^{0})$

Just like it was mentioned earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(13*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=24x+13$

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Question

What is the derivative of $2x^4+\frac{1}{2}x$ ?

Answer

To take the derivative of this equation, we can 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.

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

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

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Question

What is the derivative of ?

Answer

To take the derivative of this equation, we can 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.

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

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})+(012x^{0-1})$

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

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{0})$

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

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3$

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Question

What is the derivative of ?

Answer

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.

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

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

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Question

What is the derivative of ?

Answer

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.

For this problem that would look like this:

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

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

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

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

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