Question

x² sinx solve by derivative​

Answer 1

Question :-

$\mathcal{find \: the \: derivative \: of \: {x}^{2} \: sin \: x}$

Answer :-

$\red{\boxed{ \bf{ \frac{dy}{dx} = x \big(x \: cos \: x \: + 2 \: sin \: x)}} \: }$

Formula used :-

Here we used product rule of derivative,

$\: \boxed{\bf{\frac{d}{dx} xy = x \frac{d}{dx} y + y \frac{d}{dx}x }}$

Solution :-

$\bf{let \: \: y \: \: = {x}^{2} \: sin \: x}$

Now differentiating on both sides with respect to "x" ,by using the given formula ,

$\bf{\frac{dy}{dx} \: = {x}^{2} \frac{d}{dx} sin \: x + sin \: x \frac{d}{dx} {x}^{2} } \\ \\ \bf{ \frac{dy}{dx} = {x}^{2} \: cos \: x + sin \: x \: \times 2x} \\ \\ \bf{ \frac{dy}{dx} = {x}^{2} \: cos \: x \: + 2 x \: sin \: x} \\ \\ or \\ \boxed{ \bf{ \frac{dy}{dx} = x \big(x \: cos \: x \: + 2 \: sin \: x)}}$

Answer 2

lety=x

2

sinx

Now differentiating on both sides with respect to "x" ,by using the given formula ,

\begin{lgathered}\bf{\frac{dy}{dx} \: = {x}^{2} \frac{d}{dx} sin \: x + sin \: x \frac{d}{dx} {x}^{2} } \\ \\ \bf{ \frac{dy}{dx} = {x}^{2} \: cos \: x + sin \: x \: \times 2x} \\ \\ \bf{ \frac{dy}{dx} = {x}^{2} \: cos \: x \: + 2 x \: sin \: x} \\ \\ or \\ \boxed{ \bf{ \frac{dy}{dx} = x \big(x \: cos \: x \: + 2 \: sin \: x)}}\end{lgathered}

dx

dy

=x

2

dx

d

sinx+sinx

dx

d

x

2

dx

dy

=x

2

cosx+sinx×2x

dx

dy

=x

2

cosx+2xsinx

or

dx

dy

=x(xcosx+2sinx)