Question

differentiatie (sinx.cos^2 x + cosx. sin^2 x)​

Answer 1

Answer:

Let u=sin2x;v=cos2x

On differentiating w.r.t x respectively, we get

dxdu=2sinxcosx=sin2x

dxdv=−2cosxsinx=−sin2x

Now, dvdu=dv/dxdu/dx

=−sin2xsin2x=−1

Answer 2

Solution

$\to \sf \: \dfrac{d(sinx.cos^{2}x + cosx.sin^{2}x) }{dx}$

We Use this Properties

$\sf \to \: \dfrac{d(f(x) \pm g(x))}{dx} = \dfrac{d(f(x))}{dx} \pm \dfrac{d(g(x))}{dx}$

$\sf \to \: \dfrac{d(sinx.cos ^{2}x) }{dx} + \dfrac{d(cosx.sin^{2}x) }{dx}$

Now Using UV method

$\sf \to \: \dfrac{d(uv)}{dx} = v \bigg(\dfrac{du}{dx} \bigg) + u \bigg( \dfrac{dv}{dx} \bigg)$

Using this

$\sf \to \: cos^{2} x \dfrac{d(sinx)}{dx} + sinx \dfrac{d(cos {}^{2}x) }{dx} + sin ^{2}x \dfrac{d(cosx)}{dx} + cosx \dfrac{d(sin ^{2} x)}{dx}$

$\sf \to \: cos^{2} x \times cosx + sinx \times - 2cosxsinx + sin^{2} x \times - sinx + cosx \times 2sinxcosx$

$\sf \to \: cos ^{3} x - sinx \: sin2x - sin^{3} + cos \: sin2x$

$\sf \to \: cos ^{3} x - sin^{3} + cos \: sin2x - sinx \: sin2x$

Answer

$\sf \to \: cos ^{3} x - sin^{3} + cos \: sin2x - sinx \: sin2x$