Question

Differentiate x^x + (sinx)^x w.r.t x.

Answer 1

$let \: \: \: \: \: y = {x}^{x} + ( {sinx})^{x} \\ let \: \: \: u = {x}^{x} \: \: \: \: v = ( {sinx})^{x} \\ \\ now \: \: u = {x}^{x} \\ logu = xlogx \\ \frac{1}{u} \frac{du}{dx} = x \frac{1}{x} + logx \\ \frac{du}{dx} = u(1 + logx) \\ \frac{du}{dx} = {x}^{x} (1 + logx) \\ \\ v = ({sinx})^{x} \\ logv = xlogsinx \\ \frac{1}{v} \frac{dv}{dx} = x \frac{1}{sinx} cosx + logsinx \\ \frac{dv}{dx} = v(xcotx + logsinx) \\ \frac{dv}{dx} = ( {sinx})^{x} (xcotx + logsinx) \\ \\ \therefore \: \: \: \: \frac{dy}{dx} = {x}^{x} (1 + logx) + ( {sinx})^{x} (xcotx + logsinx)$