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High School Math : Specific Derivatives

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Example Questions

Example Question #1 : Using Implicit Differentiation

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)?

Possible Answers:

undefined

$\small \frac{-13}{54}$

$\small \frac{54}{13}$

$\small 0$

$\small \frac{-54}{13}$

Correct answer:

$\small \frac{-54}{13}$

Explanation:

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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Example Question #71 : Calculus I — Derivatives

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

Possible Answers:

$x2^{x+1}+\ln(2)x^{2}2^{x}$

$2x2^{x}-x^22^{x-1}$

$2x2^{x}+x^{3}2^{x-1}$

$2x\ln(2)2^{x}$

Correct answer:

$x2^{x+1}+\ln(2)x^{2}2^{x}$

Explanation:

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

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Example Question #1 : Specific Derivatives

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

Possible Answers:

$\frac{ \ln 3}{27}$

$-\frac{ 1}{27}$

$-\frac{ \ln 3}{27}$

$-27\ln 3$

$\frac{ 1}{27}$

Correct answer:

$-\frac{ \ln 3}{27}$

Explanation:

The instantaneous rate of change of at $x = x _{0}$ is $f ' (x _{0})$ , so we will find f ' (x) and evaluate it at x = 3 .

$\frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ 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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Example Question #73 : Calculus I — Derivatives

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

What is f'(x) ?

Possible Answers:

$f'(x) = 5 ^{x}$

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

$f'(x) = \frac{5 ^{x}}{\ln 5}$

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

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

Correct answer:

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

Explanation:

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

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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Example Question #74 : Calculus I — Derivatives

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

What is f'(4) ?

Possible Answers:

$6 \ln 6$

$\frac{\ln 6}{6}$

$36 \ln 6$

$\ln 36$

$\ln 6$

Correct answer:

$36 \ln 6$

Explanation:

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

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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Example Question #74 : Calculus I — Derivatives

Find the derivative of the following function:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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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Example Question #2 : Specific Derivatives

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

Possible Answers:

$\sec(x)$

$\csc(x)$

$\cos(x)$

$\sin(x)$

$-\sin(x)$

Correct answer:

$\cos(x)$

Explanation:

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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Example Question #2 : Specific Derivatives

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

Possible Answers:

$\sec(x)$

$\csc(x)$

$-\cos(x)$

$-\sin(x)$

$\cos(x)$

Correct answer:

$-\sin(x)$

Explanation:

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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Example Question #73 : Calculus I — Derivatives

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

Possible Answers:

$f' (x) = x^{2} \sec^{2} x^{2}$

$f' (x) = x^{2} \sec 2x$

$f' (x) = 2x \sec^{2} x^{2}$

$f' (x) = 2x \sec x^{2}$

$f' (x) = x^{2} \tan x^{2} + 2x \sec^{2} x^{2}$

Correct answer:

$f' (x) = 2x \sec^{2} x^{2}$

Explanation:

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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Example Question #5 : Specific Derivatives

Find the derivative of the following function:

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

Possible Answers:

f'(x) = 3x^2+x

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

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

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

f'(x) = 3x^2

Correct answer:

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

Explanation:

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

From the power rule, the derivative of

x^3

is simply

3x^2

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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High School Math : Specific Derivatives

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Using Implicit Differentiation

Example Question #71 : Calculus I — Derivatives

Example Question #1 : Specific Derivatives

Example Question #73 : Calculus I — Derivatives

Example Question #74 : Calculus I — Derivatives

Example Question #74 : Calculus I — Derivatives

Example Question #2 : Specific Derivatives

Example Question #2 : Specific Derivatives

Example Question #73 : Calculus I — Derivatives

Example Question #5 : Specific Derivatives

All High School Math Resources

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