Question

if y=(sinx)^tanx + (cosx)^secx , find dy/dx

Answer 1

$y = ( \sin \: x) {}^{ \tan \: x } + ( \cos \: x) {}^{ \sec\:x } \\ let \: 4 = ( \sin \: x) {}^{ \tan \: x } \\ y = u + y \: \: \: \: \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \\ u = ( \sin \: x) {}^{ \tan \: x } \: \: logu= \tan \: x \times log \: \sin \: x \\ \frac{1}{u} \times \frac{du}{dx} = \sec {}^{2} x \times log \: \sin \: x + \tan \: x \times \frac{1}{ \sin \: x \times \cos \: x } \\ \frac{du}{dx} = ( \sin \: x) {}^{ \tan \: x } [ \sec ^{2} x \times log \: \sin \: x + 1 ]$

$[ log \: \cos \: x - 1 ]$

Answer 2

Answer:

your answer attached in the photo