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AP Calculus BC : Rules of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, and Inverse Trigonometric

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Example Question #11 : Computation Of Derivatives

Find $\frac{dy}{dx}$ if $x^3 -x\ln y=ye^x$ .

Possible Answers:

$\frac{dy}{dx}=\frac{3x^2y-x-y^2e^x}{y}$

$\frac{dy}{dx}=\frac{(x^2y+y^2e^x)}{e^xy+e^x}$

$\frac{dy}{dx}=\frac{(x^2y+ln y-y^2e^x)}{e^x+x}$

cannot be determined

$\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$

Correct answer:

$\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$

Explanation:

This function is implicit, because y is not defined directly in terms of only x. We could try to solve for y, but that would be difficult, if not impossible. The easier solution would be to employ implicit differentiation. Our strategy will be to differentiate the left and right sides by x, apply the rules of differentiation (such as Chain and Product Rules), group dy/dx terms, and solve for dy/dx in terms of both x and y.

$x^3-x\ln y=ye^{x}$

$\frac{d}{dx}[x^3-x\ln y]=\frac{d}{dx}[ye^x]$

$\frac{d}{dx}[x^3]-\frac{d}{dx}[x\ln y]=\frac{d}{dx}[ye^x]$

We will need to apply both the Product Rule and Chain Rule to both the xlny and the ye^x terms.

According to the Product Rule, if f(x) and g(x) are functions, then $\frac{d}{dx}[f(x)g(x)]=f(x)\cdot \frac{d}{dx}[g(x)]+g(x)\cdot \frac{d}{dx}[f(x)]$ .

And according to the Chain Rule,

$\frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)$

$3x^2-(x\cdot \frac{d}{dx}[\ln y]+\ln y\cdot \frac{d}{dx}[x])=e^x\cdot \frac{d}{dx}[y]+y\cdot \frac{d}{dx}[e^x]$

$3x^2-(x\cdot \frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{d}{dy}[y]\cdot \frac{dy}{dx}+y\cdot e^x$

$3x^2-(\frac{x}{y}\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{dy}{dx}+ye^x$

$3x^{2}-\frac{x}y{}\cdot \frac{dy}{dx}-\ln y=e^x\cdot \frac{dy}{dx}+ye^x$

Now we will group the dy/dx terms and move everything else to the opposite side.

$3x^2-\ln y-ye^x=e^x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}=(e^x+\frac{x}{y})\frac{dy}{dx}$

Then, we can solve for dy/dx.

$\frac{3x^2-\ln y-ye^x}{e^x+\frac{x}{y}} =\frac{dy}{dx}$

To remove the compound fraction, we can multiply the top and bottom of the fraction by y.

$\frac{y(3x^2-\ln y-ye^x)}{y(e^x+\frac{x}{y})} =\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$

The (ugly) answer is $\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$ .

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Example Question #381 : Ap Calculus Bc

Find the derivative of $y= e^{2}x$ .

Possible Answers:

2e^2x

e^2x

$2e^{2x}$

e^2

None of the other answers

Correct answer:

e^2

Explanation:

This derivative uses the power rule. Keep in mind that the is not a part of the exponent of , and is thus being multiplied to e^2 . Since e^2 is a constant in front of , we have

$y' = \frac{d}{dx}(e^2x) = e^2\frac{d}{dx}(x) = e^2 \times 1 = e^2$ .

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Example Question #11 : Computation Of Derivatives

If $f(x) = 3^{2x}$ , find f'(1) .

Possible Answers:

None of the other answers

$6\ln(3)$

$2\ln(3)$

Correct answer:

$6\ln(3)$

Explanation:

In general, the derivative for an exponential function a^u is

$\ln(a) \times a^u \times \frac{du}{dx}$ .

In our case a =3 , u =2x , so we have

$f'(x) = \ln(3) \times 3^{2x} \times (2x)' = 2\ln(3) \times 3^{2x}$ .

Hence

$f'(1) = 2\ln(3) \times 3 = 6\ln{(3)}$ .

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Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Find the derivative of the following function:

$f=\ln\sec(5x)$

Possible Answers:

$\tan(5x)$

$5\sec(5x)\tan(5x)$

$5\tan(5x)$

$\cot(5x)$

$5\tan(x)$

Correct answer:

$5\tan(5x)$

Explanation:

The derivative of the function is equal to

$f'=5\tan(5x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Find the derivative of the following function:

$f(x)=\log_3(x^2)$

Possible Answers:

$\frac{2}{x\ln(3)}$

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

$\frac{2x}{\ln(3)}$

$\frac{1}{x\ln(3)}$

Correct answer:

$\frac{2}{x\ln(3)}$

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{2}{x\ln(3)}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \log_a(u)=\frac{1}{u\ln(a)}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Given h(x), find h'(x).

h(x)=3sin(x)-tan(x)+3cos(x)

Possible Answers:

h'(x)=3cos(x)-sec^2(x)-3sin(x)

h'(x)=3cos(x)+sec^2(x)-3sin(x)

h'(x)=3cos(x)-sec^2(x)+3sin(x)

h'(x)=6cos(x)-sec^2(x)+12sin(x)

Correct answer:

h'(x)=3cos(x)-sec^2(x)-3sin(x)

Explanation:

Given h(x), find h'(x).

h(x)=3sin(x)-tan(x)+3cos(x)

We need to derive a function composed of trigonometric terms.

Let's recall the rules

1) Derivative of sine is cosine

2) Derivative of cosine is negative sine

3) Derivative of tangent is secant squared.

Put this all together to get:

h(x)=3sin(x)-tan(x)+3cos(x)

h'(x)=3cos(x)-sec^2(x)-3sin(x)

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Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Compute the derivative of the function

$y=\tan^3(2x)$ .

Possible Answers:

$2\tan^2(2x)\sec(2x)$

$6\tan^2(2x)\sec^2(2x)$

$3\tan^2(2x)$

$6\tan^2(2x)$

$3\tan^2(2x)\sec^2(2x)$

Correct answer:

$6\tan^2(2x)\sec^2(2x)$

Explanation:

Although written correctly by convention, the superscript that appears immediately after the trigonometric function $\tan$ may obscure the problem; the function

$y=\tan^3(2x)$ is equivalent to writing $y=\left (\tan(2x) \right )^3$ .

Using the fact that

$\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ ,

we apply the chain rule twice, using the power rule in the first step:

$y'=3\left (\tan(2x) \right )^2\cdot\frac{\mathrm{d} }{\mathrm{d} x}\left (\tan(2x) \right )$

$=3\left (\tan(2x) \right )^2(\sec^2(2x)\cdot2)$

$=6\tan^2(2x)\sec^2(2x)$ .

(where in the last step, we have returned to the convention of writing the superscript immediately after $\tan$ )

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Example Question #118 : Derivatives

Compute the derivative of the function

$y=\ln(1+e^{-4x})$ .

Possible Answers:

$\frac{4}{e^{4x}+1}$

$\frac{1}{e^{4x}+1}$

$\frac{e^{4x}-4}{e^{4x}+1}$

$\frac{-e^{4x}}{e^{4x}-4}$

$\frac{-4}{e^{4x}+1}$

Correct answer:

$\frac{-4}{e^{4x}+1}$

Explanation:

Use the chain rule: [f(g(x))]'=f'(g(x))g'(x) with the outer function as $f(x)=\ln(x)$ and the inner function as $g(x)=1+e^{-4x}$ .

We have then:

$y'=\left (\frac{1}{1+e^{-4x}} \right )(e^{-4x}\cdot-4)$

(where we have used the chain rule again to compute the derivative of the inner function g(x) )

We can simplify this further (into the format of the answer choices) as follows:

$=\frac{-4e^{-4x}}{1+e^{-4x}}$

$=\frac{-4e^{-4x}}{1+e^{-4x}}\cdot\frac{e^{4x}}{e^{4x}}$

(multiplication by a convenient form of 1)

$=\frac{-4}{e^{4x}+1}$ .

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Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Find the first derivative of the given function.

h(t)=4sec(x)-12cot(x)+3

Possible Answers:

h'(t)=4sin(x)cot(x)-12csc^2(x)

h'(t)=4sec(x)tan(x)-12csc^2(x)

h'(t)=4sec(x)tan(x)+12csc^2(x)

$h'(t)=4csc(x)cot(x)+\frac{12}{tan(x)}$

Correct answer:

h'(t)=4sec(x)tan(x)+12csc^2(x)

Explanation:

Find the first derivative of the given function.

h(t)=4sec(x)-12cot(x)+3

So, here we need to derive a function with trigonometric terms. Let's recall the rules

1) The derivative of secant is secant times tangent

2) The derivative of cotangent is negative cosecant squared.

3) The derivative of any constant term is 0

Put all this together to get:

h(t)=4sec(x)-12cot(x)+3

h'(t)=4sec(x)tan(x)-12(-csc^2(x))

And simplify to get:

h'(t)=4sec(x)tan(x)+12csc^2(x)

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AP Calculus BC : Rules of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, and Inverse Trigonometric

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #11 : Computation Of Derivatives

Example Question #381 : Ap Calculus Bc

Example Question #11 : Computation Of Derivatives

Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Example Question #118 : Derivatives

Example Question #11 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

All AP Calculus BC Resources

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