Question

Differentiate w.r.t.x: (sin x)^x+x^sinx​

Answer 1

EXPLANATION.

$\sf \implies (sin x)^{x} \ + (x)^{sin x}.$

As we know that,

(sin x)ˣ = u

(x)^(sin x) = v.

y = u + v.

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

Taking ㏒ on both sides, we get.

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

$\sf \implies log(u) = x \ log(sinx)$

$\sf \implies \dfrac{1}{u} \times \dfrac{du}{dx} = \dfrac{d(x \ log(sinx))}{dx}$

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

$\sf \implies \dfrac{1}{u} \times \dfrac{du}{dx} = (x) \times \dfrac{1}{sinx} \times \dfrac{d(sinx)}{dx} \ + log(sin x) \times 1$

$\sf \implies \dfrac{1}{u} \times \dfrac{du}{dx} = (x) \times \dfrac{1}{sinx} \times cosx \ + log(sinx)$

$\sf \implies \dfrac{1}{u} \times \dfrac{du}{dx} = \bigg[(x)cotx \ + log(sinx)\bigg]$

$\sf \implies \dfrac{du}{dx} = u \bigg[ (x)cotx \ + log(sinx) \bigg]$

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

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

Taking ㏒ on both sides, we get.

$\sf \implies log(v) = log(x)^{sinx}$

$\sf \implies log(v) = sinx \ log(x).$

$\sf \implies \dfrac{1}{v} \times \dfrac{dv}{dx} =\dfrac{d[sinx \ log(x)]}{dx}$

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

$\sf \implies \dfrac{1}{v} \times \dfrac{dv}{dx} = sinx \times \dfrac{1}{x} \ + log(x) cosx$

$\sf \implies \dfrac{1}{v} \times \dfrac{dv}{dx} = \bigg[\dfrac{sinx}{x} \ + log(x)cos(x)\bigg]$

$\sf \implies \dfrac{dv}{dx} = v \bigg[ \dfrac{sinx}{x} \ + log(x)cos(x) \bigg]$

$\sf \implies \dfrac{dv}{dx} = (x)^{sinx}\bigg[\dfrac{sinx}{x} \ + log(x)cos(x) \bigg]$

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

$\sf \implies \dfrac{dy}{dx} = (sinx)^{x} \bigg[(x)cotx \ + log(sinx) \bigg] \ + (x)^{sinx} \bigg[ \dfrac{sinx}{x} \ + log(x)cos(x) \bigg].$