Question

Find the derivative of the function x^2cos x



Gud night!​

Answer 1

$\large \rm \red{\mp\widetilde{Answer}:-}$

Let $\rm y = x^2cosx$

Differentiating w.r.t. x both sides

$\rm :\longmapsto \dfrac{dy}{dx}=\dfrac{d}{dx}x^2cosx$

Use product rule, as:

$\underline{{\boxed{{\rm \dfrac{d}{dx} u.v = u\dfrac{dv}{dx} + v\dfrac{du}{dx}}}}}$

$\rm :\longmapsto \dfrac{dy}{dx} = x^2\dfrac{d}{dx}cosx + cosx\dfrac{d}{dx}x^2$

$\rm :\longmapsto \dfrac{dy}{dx} = x^2(-sinx) + cosx(2x^1)$

$\rm :\longmapsto \dfrac{dy}{dx} = 2xcosx -x^2sinx$

Hence,

$\bigstar \boxed{\rm \dfrac{d}{dx} x^2cosx = 2xcosx - x^2sinx}$

$\large \rm \red{\mp\widetilde{Some \ brainly\ information}:-}$

$\rm Standard \ Derivatives \\ \begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}\end{gathered}$