Question

$y=\sec x^{\tan x}+\cos x^{\sin x}$
find $\frac{dy}{dx}$

Answer 1

$y=\sec x^{\tan x}+\cos x^{\sin x}$

Let $y_1=\sec x^{\tan x},y_2=\cos x^{\sin x}$

$\therefore y_1=\sec x^{\tan x}$

taking log both side,

$\log y_1=\tan x.\log \sec x$

diff.w.r.t.x

$\frac{1}{y_1}.\frac{d{y}_1}{dx}=\tan x.\frac{1}{\sec x.}.\sec x\tan x+\log \sec x.{\sec }^{2}x$

$\frac{1}{x}.\frac{d{y}_1}{dx}={\tan }^{2}x+{\sec }^{2}x.\log secx$

$\therefore \frac{dy}{dx}=y_1.({\tan }^{2}x+{\sec }^{2}x-\log \sec x)$

$\therefore \frac{dy}{dx}\sec x^{\tan x}({\tan }^{2}x+{\sec }^{2}x.\log secx)$

$y_2=\cos x^{\sin x}$

taking log both sides,

$\log y_2=nnx.\log cosx$

diff.w.r.t.x

$\frac{1}{y_2}.\frac{d{y}_2}{dx}=\sin .\frac{1}{\cos x}.-\sin x+\log \cos x.\cos x$

$\frac{1}{y_2}.\frac{d{y}_2}{dx}=\frac{-{\sin }^{2}x}{\cos x}+\log cosx.\cos x$

$\frac{d{y}_2}{dx}=y_2.[\frac{-{\sin }^{2}x}{\cos x}+\log \cos x.\cos x]$

$\frac{d{y}_2}{dx}=\cos x^{\sin x}\left[\frac{-{\sin }^{2}x+{\cos }^{2}x.\log \cos x}{\cos x}\right]$

$\therefore \frac{dy}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$

$\therefore \frac{dy}{dx}=\sec x^{\tan x}[{\tan }^{2}x+{\sec }^{2}x.\log \sec x]+\cos x^{\sin x-1}[-{\sin }^{2}x+{\cos }^{2}x.\log x\cos x]$.