Question

Differentiate y = x square cos x?

Answer 1

Answer:

Given function : y = x² cos x.

We will use UV method to differentiate the given function.

Here, u = x² and v = cos x.

dy/dx = v du/dx + u dv/dx

➸ d(uv)/dx = cos x * d(x²)/dx + x² * d(cos x)/dx

➸ (x² cos x)' = (cos x * 2x) + (x² * -sin x)

So, differentiation of the function given is : 2x cos x - x² sin x.

RULES USED -

• y = cos x, dy/dx = sin x.

• y = xⁿ = nx^{n-1}

• UV method or Product rule of Differentiation

Answer 2

${\underline{\sf{Given}}}$

$\sf\:y=x^2\cos\:x$

${\underline{\sf{To\:Find}}}$

dy/dx

${\underline{\sf{Theory}}}$

${\red{\boxed{\large{\bold{Product\:Rule}}}}}$

Let u = f(x) and v = g(x) ,then

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

${\underline{\sf{Solution}}}$

We have,

$\sf\:y=x^2\cos\:x$

Now Differentiate with respect to x, by chain rule

$\implies\sf\dfrac{dy}{dx}=[x^2\dfrac{d(\cos\:x)}{dx}+\cos\:x\times\dfrac{d(x^2)}{dx}]$

$\implies\sf\dfrac{dy}{dx}=[-x^2\sin\:x+2x\cos\:x]$

It is the required solution!

$\rule{200}2$

${\underline{\sf{Formula's}}}$

$1)\sf\:\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$2)\sf\:\frac{d(constant)}{dx} = 0$

$3) \sf\dfrac{d(\cos\:x)}{dx} =-\sin\:x$

$\sf4) \dfrac{d(\sin\:x) }{dx} =\cos\:x$

${\red{\boxed{\large{\bold{Composite\: Function (Chain\:Rule)}}}}}$

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\sf\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$