Differentiate-sin-1-ln-x-cos-x-with-respect-to-tan-x-2-

Question Number 52006 by peter frank last updated on 01/Jan/19

D i f f e r e n t i a t e \sin^{- 1} [\frac{\ln x}{\cos x}]

w i t h r e s p e c t t o \tan x^{2}

Commented by MJS last updated on 02/Jan/19

I get (with no respect to anything)

\frac{d}{d x} [\arcsin \frac{f (x)}{g (x)}] = \frac{f^{'} g - {f g}^{'}}{g \sqrt{g^{2} - f^{2}}} =

= \frac{\cos x + x \ln x \sin x}{x \cos x \sqrt{{(\cos x)}^{2} - {(\ln x)}^{2}}}

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19

u = {s i n}^{- 1} (\frac{l n x}{c o s x})

v = {t a n x}^{2}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}}

u = {s i n}^{- 1} (\frac{l n x}{c o s x})

\frac{d u}{d x} = \frac{1}{\sqrt{1 - {(\frac{l n x}{c o s x})}^{2}}} \times \frac{c o s x (\frac{1}{x}) - l n x (- s i n x)}{{(c o s x)}^{2}}

= \frac{d u}{d x} = \frac{1}{\sqrt{1 - {(\frac{l n x}{c o s x})}^{2}}} \times \frac{\frac{c o s x}{x} + l n x (s i n x)}{{(c o s x)}^{2}}

v = {t a n x}^{2}

\frac{d v}{d x} = {s e c}^{2} (x^{2}) \times 2 x

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = [\frac{1}{\sqrt{1 - {(\frac{l n x}{c o s x})}^{2}}} \times \frac{\frac{c o s x}{x} + s i n x \times l n x}{{(c o s x)}^{2}}] \times \frac{1}{2 {x s e c}^{2} (x^{2})}

Commented by peter frank last updated on 02/Jan/19

t h a n k y o u s i r .