prove-the-formulas-given-in-table-6-for-the-derivatives-of-the-following-functions-a-cosh-1-b-tanh-1

Question

Prove the formulas given in Table 6 for the derivatives of the following functions. 
(a) $ \cosh^{-1} $ 	
(b) $ 	anh^{-1} $ 	
(c) $ csch^{-1} $
(d) $ sech^{-1} $	
(e) $ \coth^{-1} $

Clarissa Noh · Accepted Answer

he It&#x27;s clear it&#x27;s only me right here So that inverse of co sign for my public function. It is equal to L on of X plus square root of X square minus one, and for the derivative L N of X plus the square root of X square minus one, you have it equal to one over X plus square. Root that square minus one tons de over D x of X plus square root of X square minus one, which is equal to one over X plus square root specs Square minus one win won&#x27;t supply square root of X square minus one waas X over square of X square minus one, which is equal to one over square root of X square minus one part B where you make why we go up to the inverse tangent. The hyperbolic function so tangent of why is equal to X. We differentiate to get seeking square of nuts. So why d Why is equal to D. X? So we get D y o ver de acts to be won over the sequence square. So do you. Why, over DX is equal to one over one minus and Tangent Square which is equal to one over one minus expert for part c. Well, then allow Why to be able to cool seeker in verse when we get coast seeking is equal to X. We differentiate implicitly some negative co sika. Why co tangent public function of why do you Why is equal to d X? So we get you on a value for d y o ver de acts. We just stopped it too, which becomes equal to negative one all over co sika wide plus or minus squaring of one plus co Siegen square of one which is equal to a negative one over X times plus or minus square root of one plus X square which is equal to negative one over absolute value of X square root of one plus X squared. For a party, we allow why to be going to seek int in verse. So seeking of why is equal to X. So we got a deal over D X to be equal to negative one over a secret over tangent. We&#x27;re just gonna substitute and just the hyperbaric function. Why? For plus or minus square root of one minus see good square. So we got negative one over my public function of sequined, plus or minus square root one minus sequence square of why and this becomes equal to negative one over X plus or minus. It&#x27;s where it of one minus X square. So looking at the domain, we could just use the positive route since it zero comma negative zero comma, one included.

Problem

Find the derivative. Simplify where possible. $ …

Problem 29 Medium Difficulty

Prove the formulas given in Table 6 for the derivatives of the following functions.
(a) $ \cosh^{-1} $
(b) $ \tanh^{-1} $
(c) $ csch^{-1} $
(d) $ sech^{-1} $
(e) $ \coth^{-1} $

Answer

A. $\frac{1}{\sqrt{x^{2}-1}}$
B. $\frac{1}{1-x^{2}}$
C. $-\frac{1}{|x| \sqrt{x^{2}+1}}$
D. $-\frac{1}{x \sqrt{1-x^{2}}}$
E. $\frac{1}{1-x^{2}}$

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Video Transcript

Calculus: Early Transcendentals

Heather Z.

Caleb E.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

A. $\frac{1}{\sqrt{x^{2}-1}}$B. $\frac{1}{1-x^{2}}$C. $-\frac{1}{|x| \sqrt{x^{2}+1}}$D. $-\frac{1}{x \sqrt{1-x^{2}}}$E. $\frac{1}{1-x^{2}}$

Calculus: Early Transcendentals

Heather Z.

Caleb E.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Heather Z.

Caleb E.

Samuel H.

Michael J.

A. $\frac{1}{\sqrt{x^{2}-1}}$
B. $\frac{1}{1-x^{2}}$
C. $-\frac{1}{|x| \sqrt{x^{2}+1}}$
D. $-\frac{1}{x \sqrt{1-x^{2}}}$
E. $\frac{1}{1-x^{2}}$