Question

$\sec( \tan^{ - 1}x )$

Chapter: Differentiation
Standard: 12th

Please don't give useless answer. It's a Request.

PLEASE SOLVE IT . AND TELL ME THE ANSWER .

Answer 1

Here is your answer:
Let for some angle y,
$\tan( y) = x \\ \implies \: { \tan }^{ - 1} x = y$
Therefore,
$\sec({ \tan }^{ - 1} x) = \sec(y) \\ \implies \: \sec({ \tan }^{ - 1} x) = \sqrt{ { \tan }^{2}y + 1}$
But,
tan²y = x,
so,
$\sec({ \tan }^{ - 1} x) = \sqrt{ {x}^{2} + 1}$
Differentiate this thus simplified,
$\frac{d}{dx} \sec({ \tan }^{ - 1} x) = \frac{d}{dx} \sqrt{ {x}^{2} + 1}$
$\frac{d}{dx} ({{x}^{2} + 1})^{ \frac{1}{2} } \\apply \: power \: rule \\ = \frac{1}{2} ( {x}^{2} + 1)^{ \frac{1}{2} - 1} \: \times \frac{d}{dx}( {x}^{2} + 1) \\ = \frac{\frac{d}{dx}( {x}^{2} + 1)}{2( {x}^{2} + 1)} \\ = \frac{\frac{d}{dx}( {x}^{2}) +\frac{d}{dx}(1) }{2( {x}^{2} + 1)} \\ = \frac{2x +0 }{2( {x}^{2} + 1)} \\ = \frac{2x}{2( {x}^{2} + 1)} \\ \\ = \boxed{ \frac{x}{( {x}^{2} + 1)}}$
Hope you got help from my answer.☺️

Answer 2

this is my answer is it clear