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Chapter 2: Q 39. (page 237)

Fill in the blank:
$\frac{d}{d x} (\tan h^{- 1} x) =___________.$

Short Answer

Expert verified

$\frac{d}{d x} (\tan h^{- 1} x) = \frac{1}{1 - x^{2}} .$

Step by step solution

01

Given Information

The given expression is $\frac{d}{d x} (\tan h^{- 1} x) .$

02

Simplification

We know that,

\tan h^{- 1} x = \frac{1}{2} {\ln (x + 1) - \ln (1 - x)} .

we compute the derivative of $\frac{d}{d x} (\tan h^{- 1} x) .$

We get,

$\frac{d}{d x} (\tan h^{- 1} x) = \frac{d}{d x} [\frac{1}{2} {\ln (x + 1) - \ln (1 - x)} = \frac{1}{2} {\frac{1}{(x + 1)} . 1 - \frac{1}{(1 - x)} . - 1} = \frac{1}{2} (\frac{1}{x + 1} + \frac{1}{1 - x}) = \frac{1}{2} . \frac{2}{(1 - x^{2})} = \frac{1}{1 - x^{2}} T h e r e f o r e, \frac{d}{d x} (\tan h^{- 1} x) = \frac{1}{1 - x^{2}} .$

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Most popular questions from this chapter

Prove that if f is any cubic polynomial function $f (x) = a x^{3} + b x^{2} + c x + d$ then the coefficients of f are completely determined by the values of f(x) and its derivative at x=0 as follows

$a = \frac{f "' (0)}{6}; b = \frac{f " (0)}{2}; c = f' (0); d = f (0)$

Find a function that has the given derivative and value. In each case you can find the answer with an educated guess and check process it may be helpful to do some preliminary algebra

$f' (x) = {(3 x + 1)}^{3}, f (2) = 1$

Each graph in Exercises 31–34 can be thought of as the associated slope function f' for some unknown function f. In each case sketch a possible graph of f.

State the chain rule for differentiating a composition $g (h (x))$ of two functions expressed

(a) in “prime” notation and

(b) in Leibniz notation.

Use the differentiation rules developed in this section to find

the derivatives of the functions

$f (x) = 4 - x^{7}$

See all solutions

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