Question

$\huge\red{\boxed{\sf QuEsTioN}}$

Differentiate x²sin x + cos 2 x.​

Answer 1

Answer:

From the given

y

=

sin

2

x

+

cos

2

x

The right side of the equation is

=

1

y

=

1

d

y

d

x

=

d

d

x

(

1

)

=

0

or we can do it this way.

y

=

sin

2

x

+

cos

2

x

d

y

d

x

=

2

⋅

(

sin

x

)

2

−

1

⋅

d

d

x

(

sin

x

)

+

2

(

cos

x

)

2

−

1

d

d

x

(

cos

x

)

d

y

d

x

=

2

⋅

sin

x

⋅

cos

x

+

2

⋅

cos

x

⋅

(

−

sin

x

)

d

y

d

x

=

2

⋅

sin

x

⋅

cos

x

−

2

⋅

sin

x

⋅

cos

x

d

y

d

x

=

0

God bless....I hope the explanation is usef

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: {x}^{2}sinx + cos2x$

Let assume that

$\rm :\longmapsto\: y = {x}^{2}sinx + cos2x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}y =\dfrac{d}{dx}\bigg[ {x}^{2}sinx + cos2x\bigg]$

We know,

$\boxed{ \rm \:\dfrac{d}{dx}(u + v) =\dfrac{d}{dx}u + \dfrac{d}{dx}v}$

So, using this, we get

$\rm :\longmapsto\: \dfrac{dy}{dx} =\dfrac{d}{dx}{x}^{2}sinx + \dfrac{d}{dx}cos2x$

We know,

$\boxed{ \rm \:\dfrac{d}{dx}uv =u\dfrac{d}{dx}v+v\dfrac{d}{dx}u}$

and

$\boxed{ \rm \:\dfrac{d}{dx}cosx = - \: sinx}$

So, using these, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{2}\dfrac{d}{dx}sinx + sinx\dfrac{d}{dx} {x}^{2} - sin2x\dfrac{d}{dx}2x$

We know,

$\boxed{ \rm \:\dfrac{d}{dx}sinx = cosx}$

and

$\boxed{ \rm \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{2}cosx + sinx(2x) - sin2x(2)$

$\rm :\longmapsto\:\boxed{ \bf \:\dfrac{dy}{dx} = {x}^{2}cosx + 2xsinx - 2sin2x}$

Additional Information :-

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