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Calculus AB : Calculus AB

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Example Questions

Example Question #2 : Differentiate Inverse Trig Functions

Let $f(x)=cot^{-1}(4x)$ . Evaluate f'(2) .

Possible Answers:

$\frac{4}{65}$

$-\frac{4}{65}$

$-\frac{1}{65}$

$\frac{1}{65}$

Correct answer:

$-\frac{4}{65}$

Explanation:

$f(x)=cot^{-1}(4x)$

First, take the derivative of the function

$f'(x)= -\frac{4}{1+(4x)^2}= -\frac{4}{1+16x^2}$

Especially when given inverse trigonometry derivative questions, be on the lookout for multiple functions embedded in the same problem. For example, in this problem there is both an outer function ( $cot^{-1}$ ) and an inner function (). Because there is more than one function, chain rule must be applied; thus, the derivative of the inner function must be multiplied to the derivative of the outer function.

To reach the final answer, f'(x) must be evaluated at x=2 .

$f'(2)= -\frac{4}{1+16(2)^2}= -\frac{4}{65}$

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Example Question #2 : Differentiate Inverse Trig Functions

Evaluate the following derivative: $\frac{d}{dx}(sin^{-1}(3x^2))$

Possible Answers:

$\frac{x}{\sqrt{1-9x^4}}$

$\frac{1}{\sqrt{1-3x^2}}$

$\frac{6x}{\sqrt{1-9x^4}}$

$\frac{6x}{\sqrt{1-9x^2}}$

Correct answer:

$\frac{6x}{\sqrt{1-9x^4}}$

Explanation:

$\frac{d}{dx}(sin^{-1}(3x^2))=(\frac{1}{(\sqrt{1-3x^2)^2}})(6x)$

This problem requires chain rule; there is an outer function ( $sin^{-1}$ ) and an inner function ( 3x^2 ).

To simplify it further, square the 3x^2 and multiply the .

$\frac{d}{dx}(sin^{-1}(3x^2))=(\frac{1}{(\sqrt{1-3x^2)^2}})(6x) = \frac{6x}{\sqrt{1-9x^4}}$

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Example Question #2 : Differentiate Inverse Trig Functions

Evaluate the following derivative: $\frac{d}{dx}(6cos^{-1}(2x))$

Possible Answers:

$\frac{-12}{\sqrt{1-x^2}}$

$\frac{-12}{\sqrt{1-4x^2}}$

$\frac{-12}{\sqrt{1-2x^2}}$

$\frac{12}{\sqrt{1-4x^2}}$

Correct answer:

$\frac{-12}{\sqrt{1-4x^2}}$

Explanation:

$\frac{d}{dx}(6cos^{-1}(2x))=(6)(2)(-\frac{1}{\sqrt{1-(2x)^2}})$

This problem requires chain rule, because there is an outer and inner function.

The outer function is $cos^{-1}$ and the inner is .

To fully simplify the expression, multiply the constants and square the term.

$\frac{d}{dx}(6cos^{-1}(2x))=(6)(2)(-\frac{1}{\sqrt{1-(2x)^2}}) = \frac{-12}{\sqrt{1-4x^2}}$

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Example Question #3 : Differentiate Inverse Trig Functions

Let $f(x)=(2x)tan^{-1}(x)$ . What is f'(x) ?

Possible Answers:

$f'(x)=2tan^{-1}(x)$

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

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

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

Correct answer:

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

Explanation:

$f(x)=(2x)tan^{-1}(x)$

This is an example of when product rule must be used to find the derivative, because there are two functions ( and $tan^{-1}(x))$ being multiplied together within f(x) .

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

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Example Question #3 : Differentiate Inverse Trig Functions

Evaluate the following derivative: $\frac{d}{dx}(csc^{-1}(10x))$

Chain rule must be used to properly find the derivative, because there is an outer function of $csc^{-1}$ and an inner function of 10x present.

$\frac{d}{dx}(csc^{-1}(10x))=(10)(-\frac{1}{|10x| \sqrt{(10x)}^2-1})=\frac{-10}{|10x| \sqrt{100x^2-1}}$

To simplify further, the 10 in the numerator cancels out the 10 in the denominator.

$\frac{d}{dx}(csc^{-1}(10x))=\frac{-10}{|10x| \sqrt{100x^2-1}}=\frac{-1}{|x| \sqrt{100x^2-1}}$

Possible Answers:

$\frac{1}{|x| \sqrt{100x^2-1}}$

$\frac{-10}{|x| \sqrt{x^2-1}}$

$\frac{-1}{|x| \sqrt{100x^2-1}}$

$\frac{-1}{|10x| \sqrt{100x^2-1}}$

Correct answer:

$\frac{-1}{|10x| \sqrt{100x^2-1}}$

Explanation:

$\frac{d}{dx}(csc^{-1}(10x))$

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Example Question #1 : Differentiate Inverse Trig Functions

Which derivative would produce the same expression as $\frac{d}{dx}(-sec^{-1}(5x))$ ?

Possible Answers:

$\frac{d}{dx}(sec^{-1}(5x))$

$\frac{d}{dx}(-tan^{-1}(5x))$

$\frac{d}{dx}(cot^{-1}(5x))$

$\frac{d}{dx}(csc^{-1}(5x))$

Correct answer:

$\frac{d}{dx}(csc^{-1}(5x))$

Explanation:

By examining the inverse trigonometric function derivatives, it can be shown that $\frac{d}{dx}(csc^{-1}x)=\frac{d}{dx}(-sec^{-1}x)$ . The derivatives of these two inverse functions are negatives of one another.

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Example Question #1 : Differentiate Inverse Trig Functions

Let $f(x)=(sin^{-1}x)^3$ . Find f'(x) .

Possible Answers:

$\frac{cos^{-1}x}{\sqrt{1-x^2}}$

$\frac{3(sin^{-1}x)}{\sqrt{1-x^2}}$

$\frac{3(sin^{-1}x)^2}{\sqrt{1-x^2}}$

$\frac{3(cos^{-1}x)^2}{\sqrt{1-x^2}}$

Correct answer:

$\frac{3(sin^{-1}x)^2}{\sqrt{1-x^2}}$

Explanation:

$f(x)=(sin^{-1}x)^3$

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

This question invokes the use of chain rule.

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Example Question #1 : Differentiate Inverse Trig Functions

Let $f(x)=cos^{-1}(\sqrt{3x-1})$ . Find f'(x) .

Possible Answers:

$\frac{3}{2(\sqrt{3x-1})(\sqrt{2-3x})}$

$\frac{1}{(\sqrt{2-3x})}$

$\frac{-3}{2(\sqrt{3x-1})(\sqrt{2-3x})}$

$\frac{-3 sin ^{-1}x}{2(\sqrt{2-3x})}$

Correct answer:

$\frac{-3}{2(\sqrt{3x-1})(\sqrt{2-3x})}$

Explanation:

$f(x)=cos^{-1}(\sqrt{3x-1})$

This question requires the use of chain rule. In this case, there are three functions to consider: first, $cos^{-1}$ ; second, the square root; and third, 3x-1 . In order to evaluate, the derivative of each of the three functions must be found and multiplied together.

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

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Example Question #1 : Differentiate Inverse Trig Functions

Evaluate the following derivative: $\frac{d}{dx}(tan^{-1}(e^{2x}))$

Possible Answers:

$\frac{2e^{2x}}{1+e^{2x}}$

$\frac{e^{2x}}{1+e^{2x}}$

$\frac{2e^{2x}}{1+e^{4x}}$

$\frac{2}{1+e^{4x}}$

Correct answer:

$\frac{2e^{2x}}{1+e^{4x}}$

Explanation:

This problem requires use of the chain rule twice; the first time addresses the inner function of $e^{2x}$ while the second accounts for the function .

$\frac{d}{dx}(tan^{-1}(e^{2x}))=\frac{1}{1+(e^{2x})^2}*e^{2x}*2=\frac{2e^{2x}}{1+e^{4x}}$

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Example Question #1 : Implicit Differentiation And Chain Rule

$f(x)=6^(^5^x^{^{2}-7x}^)$ . Find the derivative.

Possible Answers:

f'(x)=ln6*(10x-7)

$f'(x)=6^(^5^{x^{2}-7x)}*(10x-7)*ln6$

f'(x)=(30x^2-42x)*ln6

$f'(x)=6^(^5^{x^{2}-7x)}*(10x-7)$

f'(x)=6^(^1^0^x^-^7^)*ln6

Correct answer:

$f'(x)=6^(^5^{x^{2}-7x)}*(10x-7)*ln6$

Explanation:

When the function is a constant to the power of a function of x, the first step in chain rule is to rewrite f(x). So, the first factor of f(x) will be $(6^{5x^2-7x})$ . Next, we have to take the derivative of the function that is the exponent, or 5x^2-7x . Its derivative is 10x-7, so that is the next factor of our derivative. Last, when a constant is the base of an exponential function, we must always take the natural log of that number in our derivative. So, our final factor will be ln6 . Thus, the derivative of the entire function will be all these factors multiplied together: $f'(x)=6^{5x^2-7x}*(10x-7)*ln6}$ .

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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #2 : Differentiate Inverse Trig Functions

Example Question #2 : Differentiate Inverse Trig Functions

Example Question #2 : Differentiate Inverse Trig Functions

Example Question #3 : Differentiate Inverse Trig Functions

Example Question #3 : Differentiate Inverse Trig Functions

Example Question #1 : Differentiate Inverse Trig Functions

Example Question #1 : Differentiate Inverse Trig Functions

Example Question #1 : Differentiate Inverse Trig Functions

Example Question #1 : Differentiate Inverse Trig Functions

Example Question #1 : Implicit Differentiation And Chain Rule

All Calculus AB Resources

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