Question

Find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$
$y=x\cos x$

Answer 1

Hint:If u and v are two differentiable functions of x then $\dfrac{d}{dx}\left( uv \right)=u\times \dfrac{dv}{dx}+v\times \dfrac{du}{dx}$and this formula is called product rule. In this problem u and v are two differentiable functions of x so apply the product rule. Here they asked us to find the $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$so we have to apply the product rule two times then we will get the required answer.

Complete step-by-step answer:
Given that $y=x\cos x$
We have to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$
We know that the formula for $\dfrac{d}{dx}\left( uv \right)$is given by $\dfrac{d}{dx}\left( uv \right)=v\times \dfrac{du}{dx}+u\times \dfrac{dv}{dx}$
Now apply the above formula we will get,
$\dfrac{dy}{dx}=x\left( -\sin x \right)+\cos x(1)$
$\dfrac{dy}{dx}=-x\sin x+\cos x$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Now again apply derivative to it then we will get the second derivative or double derivative.
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -x\sin x+\cos x \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -x\sin x \right)+\dfrac{d}{dx}\left( \cos x \right)$. . . . . . . . . . . . . . . . . . . . . . (3)
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-x(\cos x)+\sin x(-1)-\sin x$. . . . . . . . . . . . . . . . . . . . .(4)
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-x\cos x-2\sin x$

Note:In the above problem in the equation (3) we have used the formula that is for $\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}$and this formula is also called as sum rule. The derivative of $\sin x$is $\cos x$and the derivative of $\cos x$is $-\sin x$. So we should be keen on basic trigonometric formulas and their derivative formulas for doing this problem. Carefully do the basic mathematical operations like addition, subtraction then we will get the required answer.