# Tan √ X - Mathematics

$\tan \sqrt{x}$

#### Solution

$text{ Let } f(x) = \tan x^2$
$\text{ Thus, we have }:$
$f(x + h) = \tan (x + h )^2$
$\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{\tan (x + h )^2 - \tan x^2}{h}$
$= \lim_{h \to 0} \frac{\sin\left( \left( x + h \right)^2 - x^2 \right)}{h \cos \left( x + h \right)^2 \cos x^2} \left[ \because \tan A - \tan B = \frac{\sin (A - B)}{\cos A \cos B} \right]$
$= \lim_{h \to 0} \frac{\sin( x^2 + h^2 + 2hx - x^2 )}{h\cos \left( x + h \right)^2 \cos x^2}$
$= \lim_{h \to 0} \frac{\sin(h\left( h + 2x) \right)}{h\left( h + 2x \right) \cos \left( x + h \right)^2 \cos x^2} \times \left( h + 2x \right)$
$= \lim_{h \to 0} \frac{\sin(h\left( h + 2x) \right)}{(h\left( h + 2x) \right)} \lim_{h \to 0} \frac{h + 2x}{\cos(x + h )^2 \cos x^2} \left[ As \lim_{h \to 0} \frac{\sin(h\left( h + 2x) \right)}{(h\left( h + 2x) \right)} = 1 \right]$
$= 1 \times \frac{2x}{\cos^2 x^2}$
$= 2x \sec^2 x^2$


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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.2 | Q 5.4 | Page 26