Question

y = (sinx)^tanx + (cosx)^secx​

Answer 1

EXPLANATION.

$\sf \implies y = (sin x)^{tan x} + (cos x)^{sec x}$

As we know that,

Let we assume that,

$\sf \implies u = (sin x)^{tan x}$

$\sf \implies v = (cos x)^{sec x}$

$\sf \implies y = u + v$

$\sf \implies \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx}$

$\sf \implies u = (sin x)^{tan x}$

Taking log on both sides, we get.

$\sf \implies log (u) = log (sin x)^{tan x}$

$\sf \implies log(u) = (tan x) log (sin x)$

Differentiate both sides w.r.t x, we get.

$\sf \implies \dfrac{1}{u} . \dfrac{du}{dx} = \dfrac{d[ (tan x) log (sin x)]}{dx}$

$\sf \implies \dfrac{1}{u} . \dfrac{du}{dx} = (tan x) . \dfrac{d[log (sin x)]}{dx} \ + log (sin x) .\dfrac{d(tan x)}{dx}$

$\sf \implies \dfrac{1}{u} . \dfrac{du}{dx} = \dfrac{(tan x)}{(sin x)} . \dfrac{d(sin x)}{dx} \ + log(sin x) .(sec^{2} x)$

$\sf \implies \dfrac{1}{u} . \dfrac{du}{dx} = \bigg[ 1 + log(sin x).(sec^{2}x) \bigg]$

$\sf \implies \dfrac{du}{dx} = u \bigg[ 1 + log(sinx).(sec^{2} x) \bigg]$

$\sf \implies \dfrac{du}{dx} = (sin x)^{tan x} \bigg[ 1 + log(sin x).(sec^{2} x) \bigg].$

$\sf \implies v = (cos x)^{sec x}$

Taking log on both sides, we get.

$\sf \implies log(v) = log(cosx)^{secx}$

$\sf \implies log(v) = (sec x) log(cosx)$

Differentiate both sides w.r.t x, we get.

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = \dfrac{d[(sec x) log(cos x)]}{dx}$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = (sec x). \dfrac{d[log (cosx)]}{dx} \ + log(cosx) .\dfrac{d[secx]}{dx}$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = (sec x) \times \dfrac{1}{(cos x)} . \dfrac{d[cos x]}{dx} \ + log(cos x) . (sec x. tan x)$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = (sec x) \times \dfrac{(-sin x)}{(cos x)} \ + log(cos x).(sec x . tan x)$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = \bigg[ (sec x).(-tan x) \ + log(cos x).(sec x. tan x) \bigg]$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = \bigg[ log(cos x)(sec x .tan x) - (sec x)(tan x) \bigg]$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = \bigg[ (sec x . tan x) \bigg( log(cos x) - 1 \bigg) \bigg]$

$\sf \implies \dfrac{dv}{dx} = v \bigg[ (sec x . tan x ) \bigg( log(cos x) - 1 \bigg) \bigg]$

$\sf \implies \dfrac{dv}{dx} = (cos x)^{sec x} \bigg[ (sec x . tan x) \bigg( log(cos x) - 1 \bigg) \bigg]$

$\sf \implies y = u + v.$

$\sf \implies \dfrac{dy}{dx} = (sin x)^{tan x} \bigg[ 1 + log(sin x).(sec^{2} x) \bigg] \ + (cos x)^{sec x} \bigg[ (sec x . tan x ) \bigg( log(cos x) - 1 \bigg) \bigg]$