Question

differentiate x^sinx + sinx^cosx with respect to x​

Answer 1

$let \: \: \: \: u = {x}^{sinx} \\ \: \: \: \: \: \: \: \: v = {sinx}^{cosx} \\ \\ let \: \: \: \: y = u + v \\ \\ \frac{dy}{dx} = \frac{du}{dx} = \frac{dv}{dx} ...(1) \\ \\ now \: \: \: \: u = {x}^{sinx} \\ logu = sinxlogx \\ \frac{1}{u} \frac{du}{dx} = sinx \frac{1}{x} + logxcosx \\ \frac{du}{dx} = u( \frac{sinx}{x} + logxcosx) \\ \frac{du}{dx} = {x}^{sinx} ( \frac{sinx}{x} + logxcosx) \\ \\ v = {sinx}^{cosx} \\ logv = cosxlogsinx \\ \frac{1}{v} \frac{dv}{dx} = cosx \frac{1}{sinx} cosx + logsinx( - sinx) \\ \frac{dv}{dx} = v( \frac{ {cos}^{2} x}{sinx} - sinxlogsinx) \\ \frac{dv}{dx} = {sinx}^{cosx} ( \frac{ {cos}^{2}x }{sinx} - sinxlogsinx) \\ \\ \\ \frac{dy}{dx} = {x}^{sinx} ( \frac{sinx}{x} + logxcosx) + {sinx}^{cosx} ( \frac{ {cos}^{2}x }{sinx} - sinxlogsinx)$