Question

derivative of sin^2x^3.cos^3x^2​

Answer 1

$y = {sin}^{2} {x}^{3} . {cos}^{3} {x}^{2} \\ let \: \: \: u = {sin}^{2} {x}^{3} \: \: and \: \: v = {cos}^{3} {x}^{2} \\ y = uv \\ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}...(1) \\ finding \: derivative \: of \: u \\ u = {sin}^{2} {x}^{3} \\ \frac{du}{dx} = 2sin {x}^{3} (cos {x}^{3} )3 {x}^{2} \\ = 6 {x}^{2} sin {x}^{3} \: cos {x}^{3} \\ finding \: derivative \: of \: v \\ v = {cos}^{3} {x}^{2} \\ \frac{dv}{dx} = 3 {cos}^{2} {x}^{3} ( - sin {x}^{2} )2x \\ = - 6x {cos}^{2} {x}^{3} sin {x}^{2} \\ now \: \: from \: \: eqn \: \: (1) \\ \frac{dy}{dx} = {sin}^{2} {x}^{3} ( - 6x{sinx}^{2} {cos}^{2} {x}^{3} ) + {cos}^{3} {x}^{2} (6 {x}^{2} sin {x}^{3} cos {x}^{3} )$