Specific Derivatives - High School Math

Card 0 of 14

Question

Find the derivative $y^{'}$ for $y = x^{2}2^x$

Answer

The derivative must be computed using the product rule. Because the derivative of brings a down as a coefficient, it can be combined with to give $2^{x+1}$

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Question

$f(x) = 5 ^{x + 1}$

What is ?

Answer

$f(x) = 5 ^{x + 1} = 5 ^{1} \cdot 5 ^{x } = 5 \cdot 5 ^{x }$

Therefore,

$f'(x) = 5 \cdot \frac{\mathrm{d} }{\mathrm{d} x} \left ( 5 ^{x } \right )$

$\frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any real , so $\frac{\mathrm{d} }{\mathrm{d} x} \left ( 5 ^{x } \right ) = \ln 5 \cdot 5 ^{x}$ , and

$f'(x) = 5 \ln 5 \cdot 5 ^{x} = \ln 5 \cdot 5 \cdot 5 ^{x} = \ln 5 \cdot 5 ^{x+1}$

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Question

$f(x) = 6 ^{x - 2}$

What is ?

Answer

$f(x) = 6 ^{x -2} = 6 ^{-2} \cdot 6 ^{x } = \frac{1}{36} \cdot 6 ^{x }$

Therefore,

$f'(x) = \frac{1}{36} \cdot \frac{\mathrm{d} }{\mathrm{d} x} \left ( 6 ^{x } \right )$

$\frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so $\frac{\mathrm{d} }{\mathrm{d} x} \left ( 6 ^{x } \right ) = \ln 6 \cdot 6 ^{x}$ , and

$f'(x) = \frac{1}{36} \ln 6 \cdot 6 ^{x} = \ln 6 \cdot \frac{1}{36} \cdot 6 ^{x}$

$f'(x) = \ln 6 \cdot 6 ^{x-2}$

$f'(4) = \ln 6 \cdot 6 ^{4-2} = \ln 6 \cdot 6 ^{2} = 36 \ln 6$

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Question

Give the instantaneous rate of change of the function $f (x) = \left ( \frac{1}{3} \right ) ^{x}$ at .

Answer

The instantaneous rate of change of at $x = x _{0}$ is $f ' (x _{0})$ , so we will find and evaluate it at .

$\frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so

$f'(x) =\frac{\mathrm{d}}{\mathrm{d} x}\left [ \left ( \frac{1}{3} \right ) ^{x }\right ] = \ln \frac{1}{3} \cdot \left ( \frac{1}{3} \right) ^{x }= -\frac{ \ln 3}{3^{x}}$

$f'(3) =-\frac{ \ln 3}{3^{3}} = -\frac{ \ln 3}{27}$

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Question

Find the derivative of the following function:

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

Answer

Since this function is a polynomial, we take the derivative of each term separately.

From the power rule, the derivative of

is simply

We can rewrite $\frac{1}{x}$ as

$x^{-1}$

and using the power rule again, we get a derivative of

$-x^{-2}$ or $-\frac{1}{x^2}$

So the answer is

$f'(x) = 3x^2-\frac{1}{x^2}$

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Question

Which of the following best represents $\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$ ?

Answer

The question is just asking for the Quotient Rule formula.

Recall the Quotient Rule is the bottom function times the derivative of the top minus the top function times the derivative of the bottom all divided by the bottom function squared.

Given,

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$

the bottom function is and the top function is . This makes the bottom derivative and the top derivative .

Substituting these into the Quotient Rule formula resulting in the following.

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{(g(x))^2}$

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Question

What is $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} ?$

Answer

The chain rule is "first times the derivative of the second plus second times derivative of the first".

In this case, that means $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} =f(x)g'(x)+f'(x)g(x)$ .

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Question

What is the first derivative of $5x^4+\sin x$ ?

Answer

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, $\frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$ .

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(4*5x^{4-1})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$

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Question

What is the second derivative of $5x^4+\sin x$ ?

Answer

To find the second derivative, we need to start by finding the first one.

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, $\frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$ .

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(45x^{4-1})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$

Now we repeat the process, but using $20x^3+\cos x$ as our equation.

$\frac{\mathrm{d} }{\mathrm{d} x} \cos x=-\sin x$

$\frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(320x^{3-1})-\sin x$

$\frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(60x^{2})-\sin x$

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Question

Find the derivative of the following function:

$f(x) = 3 ;\textup{sin}(x)$

Answer

The derivative of $\sin (x)$ is $\cos (x)$ . It is probably best to memorize this fact (the proof follows from the difference quotient definition of a derivative).

Our function

$f(x) = 3 ;\textup{sin}(x)$

the factor of 3 does not change when we differentiate, therefore the answer is

$f(x) = 3 ;\textup{cos}(x)$

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Question

Compute the derivative of the function $f(x) = \tan x^{2}$ .

Answer

Use the Chain Rule.

Set $u = x^{2}$ and substitute.

$f(x) = \tan u$

$\frac{\mathrm{d} f }{\mathrm{d} u} = \frac{\mathrm{d} }{\mathrm{d} u} \tan u = \sec^{2} u$

$\frac{\mathrm{d} u }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} x^{2} = 2x^{2-1} =2x$

$\frac{\mathrm{d} f }{\mathrm{d} x} = \frac{\mathrm{d} f }{\mathrm{d} u} \cdot \frac{\mathrm{d} u }{\mathrm{d} x}= \sec^{2} u \cdot 2x = 2x \sec^{2} x^{2}$

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Question

$\frac{\mathrm{d}}{\mathrm{d} x}\sin(x)=?$

Answer

The derivative of a sine function does not follow the power rule. It is one that should be memorized.

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$ .

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Question

What is the second derivative of $\sin(x)$ ?

Answer

The derivatives of trig functions must be memorized. The first derivative is:

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$ .

To find the second derivative, we take the derivative of our result.

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-\sin(x)$ .

Therefore, the second derivative will be $-\sin(x)$ .

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Question

An ellipse is represented by the following equation:

$\small 8x^{2}+3y^{2}+6x+y-104 = 0$

What is the slope of the curve at the point (3,2)?

Answer

It would be difficult to differentiate this equation by isolating . Luckily, we don't have to. Use $\small \frac{dy}{dx}$ to represent the derivative of with respect to and follow the chain rule.

(Remember, $\small \frac{dx}{dx}$ is the derivative of with respect to , although it usually doesn't get written out because it is equal to 1. We'll write it out this time so you can see how implicit differentiation works.)

$\small 8x^{2}+3y^{2}+6x+y-104 = 0$

$\small 16x\frac{dx}{dx}+6y\frac{dy}{dx}+6\frac{dx}{dx}+\frac{dy}{dx}=0$

$\small 16x(1)+6y\frac{dy}{dx}+6(1)+\frac{dy}{dx}=0$

$\small 16x+6y\frac{dy}{dx}+6+\frac{dy}{dx}=0$

Now we need to isolate $\small \frac{dy}{dx}$ by first putting all of these terms on the same side:

$\small \small 6y\frac{dy}{dx}+\frac{dy}{dx}=-16x-6$

$\small \frac{dy}{dx}(6y+1)=-16x-6$

$\small \frac{dy}{dx}=\frac{-16x-6}{6y+1}$

This is the equation for the derivative at any point on the curve. By substituting in (3, 2) from the original question, we can find the slope at that particular point:

$\small \frac{dy}{dx}=\frac{-16(3)-6}{6(2)+1}=\frac{-48-6}{12+1}=\frac{-54}{13}$

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