Question

if y = (sin x)^ tanx + ( cos x )^secx find dy/dx​

Answer 1

Step-by-step explanation:

$\tt y = {sinx}^{tanx} + {cosx}^{secx} \\ \\ \tt y = u + v \\ \\ \tt u = {sinx}^{tanx} \\ \\ \tt Taking\:log\:on\:both\:sides \\ \\ \tt \longrightarrow log u = log {sinx}^{tanx} \\ \\ \tt \longrightarrow log u = tanx logsinx \\ \\ \tt Differentiating\: w.r.t \: x \\ \\ \tt \longrightarrow \frac{1}{u} \frac{du}{dx} = \frac{d(tanx)}{dx} logsinx + \frac{d(logsinx)}{dx} tanx \\ \\ \tt \longrightarrow \frac{1}{u}\frac{du}{dx} = {sec}^{2} xlogsinx + \frac{cosx}{sinx}tanx \\ \\ \tt \longrightarrow \frac{1}{u} \frac{du}{dx} = logsinx {sec}^{2}x + 1 \\ \\ \tt \ longrightarrow \frac{du}{dx} = u({sec}^{2} xlogsinx + 1 ) \\ \\ \longrightarrow \boxed{\bold{\underline{\red{\tt \frac{du}{dx} = {sinx}^{tanx}({sec}^{2} xlogsinx + 1 ) }}}} \\ \\ \tt Now \: v = {cosx}^{secx} \\ \\ \tt Taking\:log\:on\:both\:sides \\ \\ \tt \longrightarrow logv = log{cosx}^{secx} \\ \\ \tt \longrightarrow logv = secx logcosx \\ \\ \tt Differentiating \: w.r.t. \: x \\ \\ \tt \longrightarrow \frac{1}{v}\frac{dv}{dx} = \frac{d(secx)}{dx} logcosx + \frac{d(logcosx)}{dx} secx \\ \\ \tt \longrightarrow \frac{1}{v}\frac{dv}{dx} = secxtanx logcosx + \frac{sinx}{cosx}secx \\ \\ \tt \longrightarrow \frac{1}{v}\frac{dv}{dx} =secxtanx logcosx + secx tanx \\ \\ \tt \longrightarrow \frac{dv}{dx} = vsecxtanx ( logcosx + 1 ) \\ \\ \tt \longrightarrow \boxed{\bold{\underline{\red{\tt \frac{dv}{dx} = {cosx}^{secx} secxtanx ( logcosx + 1) }}}} \\ \\ \tt \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \\ \\ \tt \frac{dy}{dx} = {sinx}^{tanx}({sec}^{2} xlogsinx + 1 ) + {cosx}^{secx} secxtanx ( logcosx + 1)$

Answer 2

Answer:

your answer attached in the photo