Answer to Question #119031 in Calculus for Max

Question

Differentiate each function given below with respect to x.

(i) f(x) = √tan^−1x

(ii) f(x) = cos^−1(e^2x)

(iii) f(x) = sin^−1(√1− √x^2 )

(iv)f(x) = tan^−1(x/√x-x^2)

(v) f(x) = sin( tan^−1 4x)

Answer 1

(i) $f(x)=\sqrt{\tan^{-1}(x)}$

Differentiating with respect to x both sides,

$f'(x) = [\frac{1}{2\sqrt {tan^{-1} x}}] [\frac{1}{1 + x^2}]$

(ii) f(x) = cos^-1(e^2x)

Differentiating with respect to x both sides,

$f'(x) = [\frac{-2e^{2x}}{\sqrt{1 - {e^{4x} }}} ]$

$(iv) f(x) = tan^{-1}\frac{x}{\sqrt{x - x^{2}}}$

Differentiating with respect to x both sides,

$f'(x) = [\frac{1}{1 + {\frac{x^{2}}{x - x^{2}} }} ] [\frac{\sqrt{x - x^{2}} - \frac{x{(1 - 2x)}}{2\sqrt{x - x^{2}}}}{x - x^{2}}]$

Solving equation, we get,

$f'(x) = \frac{(1-x)(2x - 2x^{2} - x + 2x^{2})}{2 {(x - x^{2})}^{3/2}}$

$f'(x) = \frac{1}{2 } [ ({x - x^{2}})^{(- {\frac{1}{2}})}]$

$(iii) f(x) = {sin^{-1}{\sqrt{1 - \sqrt{x^{2} }}}}$

$so, f(x) will be \implies f(x) = {sin^{-1}{\sqrt{1 - x}}}$

Differentiating both sides w.r.t. x

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

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

(v) F(x) = sin(tan^-14x)

Differentiating both sides with respect to x,

$f'(x) = \frac{(4 cos(tan^{-1}4x))}{1 + 16x^{2}}$