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Chapter 2: Q 26 (page 238)

Find the derivatives of the function
$f (x) = \frac{\sin h^{- 1} x}{\tan h^{- 1} x}$

Short Answer

Expert verified

$f' (x) = \frac{d y}{d x} = \frac{1}{\sqrt{x^{2} - 1}}$

Step by step solution

01

Step 1:Given information

The given expression is $f (x) = \frac{\sin h^{- 1} x}{\tan h^{- 1} x}$

02

Step 2:Simplification

Given

$f (x) = y = \frac{\sin h^{- 1} x}{\tan h^{- 1} x}$

$\Rightarrow y = \frac{\sin h^{- 1} x}{\frac{\sin h^{- 1} x}{\cos h^{- 1} x}}$

$\Rightarrow y = \sin h^{- 1} x . \frac{\cos h^{- 1} x}{\sin h^{- 1} x}$

$\Rightarrow y = \cos h^{- 1} x$

Differentiating w.r.t x

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (\cos h^{- 1} x)$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\sqrt{x^{2} - 1}}$

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Most popular questions from this chapter

Use the definition of the derivative to find $f'$ for each function $f$ in Exercises 34-59

role="math" localid="1648284617718" $f (x) = \sqrt{2 x + 1}$

A bowling ball dropped from a height of $400$ feet will be $s (t) = 400 - 16 t^{2}$ feet from the ground after $t$ seconds. Use a sequence of average velocities to estimate the instantaneous velocities described below:

After $t = 1$ seconds, with $h = 0.5, h = 0.25, h = - 0.5 a n d h = - 0.2$

Use the definition of the derivative to find the equations of the lines described in Exercises 59-64.

The tangent line to $f (x) = x^{2}$ at $x = 0$

Use the $h \to 0$ definition of the derivative to prove the power rule holds for positive integers powers

Differentiate $f (x) = {(3 x + \sqrt{x})}^{2}$ in three ways. When you have completed all three parts, show that your three answers are the same:

(a) with the chain rule

(b) with the product rule but not the chain rule

(c) without the chain or product rules.

See all solutions

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