Question

if Y equal to sin square x cos square x then find DY by DX​

Answer 1

Answer:

$\boxed{\mathfrak{ \dfrac{d}{dx} ( {sin}^{2} x \: {cos}^{2} x) = 2sinx \: cosx \: ( {cos}^{2} x - {sin}^{2} x)}}$

Given:

$\rm y = {sin}^{2} x \: {cos}^{2} x$

Explanation:

$\rm Possible \: derivation: \\ \rm \implies \dfrac{d}{dx} ( {sin}^{2} x \: {cos}^{2} x) \\ \\ \rm Using \: the \: product \: rule, \\ \rm \dfrac{d}{dx} (u v) = v( \frac{d}{dx}u) + u ( \frac{d}{dx} v) \\ \\ \rm \implies {sin}^{2} x( \dfrac{d}{dx} {cos}^{2} x) + {cos}^{2} x( \dfrac{d}{dx} {sin}^{2} x) \\ \\ \rm Using \: the \: chain \: rule : \\ \\ \rm \implies {sin}^{2} x( \dfrac{d}{dx} {cos}^{2} x \times \dfrac{d \: cosx}{d \: cos \: x} ) + {cos}^{2} x( \dfrac{d}{dx} {sin}^{2} x\times \dfrac{d \: sinx}{d \: sin \: x}) \\ \\ \rm \implies {sin}^{2} x( \dfrac{d}{d \: cosx} {cos}^{2} x \times \dfrac{d \: cosx}{d x} ) + {cos}^{2} x( \dfrac{d}{d \: sinx} {sin}^{2} x\times \dfrac{d \: sinc}{dx}) \\ \\ \rm \implies 2cos x \: {sin}^{2} x( \dfrac{d }{d x} cosx) + 2sinx \: {cos}^{2} x( \dfrac{d }{d x}sinx) \\ \\ \rm \implies 2cos x \: {sin}^{2} x( - sinx) + 2sinx \: {cos}^{2} x( cosx) \\ \\ \rm \implies - 2cos x \: {sin}^{3} x + 2sinx \: {cos}^{3} x\\ \\ \rm \implies 2sinx \: {cos}^{3} x- 2cos x \: {sin}^{3} x \\ \\ \rm \implies 2sinx \: cosx \: ( {cos}^{2} x - {sin}^{2} x)$