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 $y = e^{6x} \cos 3x$. Using the product rule,$\frac{dy}{dx} = \frac{d}{dx}(e^{6x}) \cdot \cos 3x + e^{6x} \cdot \frac{d}{dx}(\cos 3x)$. $\frac

Vedclass · Accepted Answer

$9 e^{6x}(3 \cos 3x - 4 \sin 3x)$

Vedclass · Answer

(A) Let $y = e^{6x} \cos 3x$. Using the product rule,$\frac{dy}{dx} = \frac{d}{dx}(e^{6x}) \cdot \cos 3x + e^{6x} \cdot \frac{d}{dx}(\cos 3x)$. $\frac{dy}{dx} = 6e^{6x} \cos 3x + e^{6x} (-3 \sin 3x) = 6e^{6x} \cos 3x - 3e^{6x} \sin 3x$. Now,differentiate again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(6e^{6x} \cos 3x) - \frac{d}{dx}(3e^{6x} \sin 3x)$. $\frac{d^2y}{dx^2} = 6[6e^{6x} \cos 3x - 3e^{6x} \sin 3x] - 3[6e^{6x} \sin 3x + 3e^{6x} \cos 3x]$. $\frac{d^2y}{dx^2} = 36e^{6x} \cos 3x - 18e^{6x} \sin 3x - 18e^{6x} \sin 3x - 9e^{6x} \cos 3x$. $\frac{d^2y}{dx^2} = 27e^{6x} \cos 3x - 36e^{6x} \sin 3x$. Factoring out $9e^{6x}$,we get: $\frac{d^2y}{dx^2} = 9e^{6x}(3 \cos 3x - 4 \sin 3x)$.

Find the second order derivative of the function $y = e^{6x} \cos 3x$.

Similar Questions

If $y = \operatorname{Sin}^{-1}\left(\frac{2x}{1+x^2}\right)$ and $\left(\frac{d^2y}{dx^2}\right)_{x=2} = k$,then $25k =$

If $y = e^{nx}$,then $\left( \frac{d^2y}{dx^2} \right) \left( \frac{d^2x}{dy^2} \right)$ is equal to

For $y=\sin ^{-1}\left\{\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right\} ;|x| \leq 1$, if $a\left(1-x^{2}\right) y_{2}+b x y_{1}=0$ then $(a, b)=$

If $\sqrt{x+y}+\sqrt{y-x}=5$,then $\left(\frac{d^{2} y}{d x^{2}}\right)=$

$\frac{d^2}{dx^2}(2\cos x \cos 3x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module