Differentiation

Given, $ x \sqrt{1+y}+y \sqrt{1+x}=0 $

$ \begin{array}{ll} \Rightarrow & x \sqrt{1+y}=-y \sqrt{1+x} \\ \Rightarrow & (x \sqrt{1+y})^{2}=(-y \sqrt{1+x})^{2} \end{array} $

[squaring both sides]

$ \begin{array}{lr} \Rightarrow & x^{2}(1+y)=y^{2}(1+x) \\ \Rightarrow & x^{2}-y^{2}=y^{2} x-x^{2} y \\ \Rightarrow & (x-y)(x+y)=x y(y-x) \\ \Rightarrow & x+y=-x y \\ \Rightarrow & x+y+x y=0 \\ \Rightarrow & y=\frac{-x}{1+x}=-1+\frac{1}{1+x} \end{array} $

Differentiating w.r.t, $ x $

$ \frac{d y}{d x}=-\frac{1}{(1+x)^{2}} $

Given, $y = \sin \left( {2{{\tan }^{ - 1}}\sqrt {{{1 - x} \over {1 + x}}} } \right)$

Let $x = \cos 2\theta \Rightarrow y = \sin \left( {2{{\tan }^{ - 1}}\sqrt {{{1 - \cos 2\theta } \over {1 + \cos 2\theta }}} } \right)$

$ \Rightarrow y = \sin \left( {2{{\tan }^{ - 1}}\sqrt {{{2{{\sin }^2}\theta } \over {2{{\cos }^2}\theta }}} } \right) \Rightarrow y = \sin [2{\tan ^{ - 1}}(\tan \theta )]$

$ \Rightarrow y = \sin 2\theta \Rightarrow y = \sqrt {1 - {{\cos }^2}2\theta } \Rightarrow y = \sqrt {1 - {x^2}} $

$\therefore$ ${{dy} \over {dx}} = {1 \over {2\sqrt {1 - {x^2}} }}( - 2x) = - {x \over {\sqrt {1 - {x^2}} }}$

Let f(x) = Ax² + Bx+ C

$\therefore$ f(1) = A + B + C

and f($-$1) = A $-$ B + C

$\because$ f(1) = f($-$1)

$\Rightarrow$ A + B + C = A $-$ B + C

$\Rightarrow$ 2B = 0 $\Rightarrow$ B = 0

$\therefore$ f(x) = Ax² + C

$\Rightarrow$ f'(x) = 2Ax

$\Rightarrow$ f'(a) = 2Aa

f'(b) = 2Ab and f'(c) = 2Ac

Also, a, b, c are in AP

So, 2Aa, 2Ab and 2Ac are in AP.

$\Rightarrow$ f'(a), f'(b) and f'(c) are also in AP.

We have,

$f(x) = 2{x^3} + {x^4} + \log x$

Put f(x) = 3, we get

x = 1

$\therefore$ $f(1) = 3 \Rightarrow 1 = {f^{ - 1}}(3)$

Now, given g is the inverse of f

$\therefore$ $g(x) = {f^{ - 1}}(x) \Rightarrow f(g(x)) = x$

Differentiating with respect to x, we get

$f'(g(x))\,.\,g'(x) = 1$

$g'(x) = {1 \over {f'(g(x))}}$

$g'(x) = {1 \over {{{(g(x))}^2} + 4{{(g(x))}^3} + {1 \over {g(x)}}}}$ [$\because$ $f'(x) = 6{x^2} + 4{x^3} + {1 \over x}$]

$g'(3) = {1 \over {6{{(g(3))}^2} + 4{{(g(3))}^3} + {1 \over {g(3)}}}}$

$g'(3) = {1 \over {6 + 4 + 1}}$ [$\because$ $g(3) = {f^{ - 1}}(3) = 1$]

$g'(3) = {1 \over {11}}$