Find the second order derivative of the funct | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $y = x \cdot \cos x$. First,we find the first derivative using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \cos x + x \cdot \frac

Vedclass · Accepted Answer

$-(x \cos x + 2 \sin x)$

Vedclass · Answer

(A) Let $y = x \cdot \cos x$. First,we find the first derivative using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \cos x + x \cdot \frac{d}{dx}(\cos x)$ $= 1 \cdot \cos x + x \cdot (-\sin x) = \cos x - x \sin x$. Now,we find the second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\cos x - x \sin x) = \frac{d}{dx}(\cos x) - \frac{d}{dx}(x \sin x)$. Using the product rule for the second term: $\frac{d}{dx}(x \sin x) = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x) = 1 \cdot \sin x + x \cos x = \sin x + x \cos x$. Substituting this back: $\frac{d^2y}{dx^2} = -\sin x - (\sin x + x \cos x) = -\sin x - \sin x - x \cos x = -(x \cos x + 2 \sin x)$.

Find the second order derivative of the function $f(x) = x \cdot \cos x$.

Similar Questions

The $n^{th}$ derivative of $x^{n + 1}$ is

If $f(x) = \sin x + \cos x$, then $f\left(\frac{\pi}{4}\right) f^{(iv)}\left(\frac{\pi}{4}\right)$ is equal to

If $2x = y^{1/5} + y^{-1/5}$ and $(x^2 - 1) \frac{d^2y}{dx^2} + \lambda x \frac{dy}{dx} + ky = 0$,then $\lambda + k$ is equal to

If $\sqrt{x+y}-\sqrt{x-y}=c$, then $\frac{d^2 y}{d x^2}=$

If $e^y(x+1)=1$, then $\frac{d^2y}{dx^2} - \left(\frac{dy}{dx}\right)^2 = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module