Section 3.6: Derivatives of Inverse Functions

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Flashcards covering Section 3.6: Derivatives of Inverse Functions, including the Inverse Function Derivative Rule, derivatives of logarithms, logarithmic differentiation, the generalized power rule, and derivatives of inverse trigonometric functions.

20 Terms

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According to the Derivative Rule for Inverse Functions, what is the formula for (f−1)′(f(x))(f^{-1})'(f(x)) when x∈(a,b)x \in (a, b) and f′(x)≠0f'(x) \neq 0?

(f−1)′(f(x))=1f′(x)(f^{-1})'(f(x)) = \frac{1}{f'(x)}

2
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According to the Derivative Rule for Inverse Functions, what is the formula for (f−1)′(x)(f^{-1})'(x) for x∈f((a,b))x \in f((a, b))?

(f−1)′(x)=1f′(f−1(x))(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}

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What is the derivative of the natural logarithm function ln⁡(x)\ln(x) for all x>0x > 0?

ddx(ln⁡(x))=1x\frac{d}{dx}(\ln(x)) = \frac{1}{x}

4
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Given a positive base a≠1a \neq 1, what is the derivative of log⁡a(x)\log_a(x) for all x>0x > 0?

ddx(log⁡a(x))=1ln⁡(a)⋅1x\frac{d}{dx}(\log_a(x)) = \frac{1}{\ln(a)} \cdot \frac{1}{x}

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What are the three steps of Logarithmic Differentiation?

Step 1: Apply the natural logarithm to both sides of y=f(x)y = f(x) and simplify using log properties. Step 2: Assuming yy is differentiable, differentiate both sides implicitly with respect to xx. Step 3: Solve the resulting equation for y′y'.

6
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In the proof of the Power Rule using logarithmic differentiation for y=xry = x^r, what intermediate step is taken before taking logarithms?

∣y∣=∣x∣r|y| = |x|^r, which allows taking the natural logarithm of both sides to get ln⁡(∣y∣)=rln⁡(∣x∣)\ln(|y|) = r \ln(|x|).

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What is the Generalized Power Rule for a differentiable function u=u(x)u = u(x)?

ddx(ur)=rur−1dudx\frac{d}{dx}(u^r) = r u^{r-1} \frac{du}{dx}

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What are the domain and range of the inverse sine function y=sin⁡−1(x)y = \sin^{-1}(x)?

Domain: [−1,1][-1, 1]; Range: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]

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What are the domain and range of the inverse cosine function y=cos⁡−1(x)y = \cos^{-1}(x)?

Domain: [−1,1][-1, 1]; Range: [0,π][0, \pi]

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<p>Which inverse trigonometric function is represented in this graph, and what are its stated domain and range?</p>

Which inverse trigonometric function is represented in this graph, and what are its stated domain and range?

The inverse tangent function y=tan⁡−1(x)y = \tan^{-1}(x), with domain (−∞,∞)(-\infty, \infty) and range (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

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What is the identity that relates csc⁡−1(x)\csc^{-1}(x) to sin⁡−1\sin^{-1}?

csc⁡−1(x)=sin⁡−1(1x)\csc^{-1}(x) = \sin^{-1}\left(\frac{1}{x}\right)

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What is the identity that relates sec⁡−1(x)\sec^{-1}(x) to cos⁡−1\cos^{-1}?

sec⁡−1(x)=cos⁡−1(1x)\sec^{-1}(x) = \cos^{-1}\left(\frac{1}{x}\right)

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What identity relates cot⁡−1(x)\cot^{-1}(x) to tan⁡−1\tan^{-1}, and what is the value of cot⁡−1(0)\cot^{-1}(0)?

cot⁡−1(x)=tan⁡−1(1x)\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right), with cot⁡−1(0)=π2\cot^{-1}(0) = \frac{\pi}{2}.

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<p>In the derivation of the derivative of $$\tan^{-1}(x)$$, what are the side lengths of this reference right triangle when $$\theta = \tan^{-1}(x)$$?</p>

In the derivation of the derivative of tan⁡−1(x)\tan^{-1}(x), what are the side lengths of this reference right triangle when θ=tan⁡−1(x)\theta = \tan^{-1}(x)?

Opposite side = xx, adjacent side = 11, and hypotenuse = 1+x2\sqrt{1 + x^2}.

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What is the derivative of sin⁡−1(u)\sin^{-1}(u) with respect to xx using the chain rule?

ddx(sin⁡−1(u))=11−u2dudx\frac{d}{dx}(\sin^{-1}(u)) = \frac{1}{\sqrt{1 - u^2}} \frac{du}{dx}

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What is the derivative of cos⁡−1(u)\cos^{-1}(u) with respect to xx using the chain rule?

ddx(cos⁡−1(u))=−11−u2dudx\frac{d}{dx}(\cos^{-1}(u)) = -\frac{1}{\sqrt{1 - u^2}} \frac{du}{dx}

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What is the derivative of tan⁡−1(u)\tan^{-1}(u) with respect to xx using the chain rule?

ddx(tan⁡−1(u))=11+u2dudx\frac{d}{dx}(\tan^{-1}(u)) = \frac{1}{1 + u^2} \frac{du}{dx}

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What is the derivative of cot⁡−1(u)\cot^{-1}(u) with respect to xx using the chain rule?

ddx(cot⁡−1(u))=−11+u2dudx\frac{d}{dx}(\cot^{-1}(u)) = -\frac{1}{1 + u^2} \frac{du}{dx}

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What is the derivative of sec⁡−1(u)\sec^{-1}(u) with respect to xx using the chain rule?

ddx(sec⁡−1(u))=1∣u∣u2−1dudx\frac{d}{dx}(\sec^{-1}(u)) = \frac{1}{|u|\sqrt{u^2 - 1}} \frac{du}{dx}

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What is the derivative of csc⁡−1(u)\csc^{-1}(u) with respect to xx using the chain rule?

ddx(csc⁡−1(u))=−1∣u∣u2−1dudx\frac{d}{dx}(\csc^{-1}(u)) = -\frac{1}{|u|\sqrt{u^2 - 1}} \frac{du}{dx}