If y = 100 e^{2x} + 200 e^{-2x} and frac{d^2 | vedclass

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

Question

(D) Given the equation $y = 100 e^{2x} + 200 e^{-2x}$.First,find the first derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = 100 \cdot 2 e^{2x} + 200 \cdot

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given the equation $y = 100 e^{2x} + 200 e^{-2x}$.First,find the first derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = 100 \cdot 2 e^{2x} + 200 \cdot (-2) e^{-2x} = 200 e^{2x} - 400 e^{-2x}$.Next,find the second derivative $\frac{d^2 y}{dx^2}$:$\frac{d^2 y}{dx^2} = 200 \cdot 2 e^{2x} - 400 \cdot (-2) e^{-2x} = 400 e^{2x} + 800 e^{-2x}$.Factor out $4$ from the expression:$\frac{d^2 y}{dx^2} = 4(100 e^{2x} + 200 e^{-2x})$.Since $y = 100 e^{2x} + 200 e^{-2x}$,we can substitute $y$ into the equation:$\frac{d^2 y}{dx^2} = 4y$.Comparing this with $\frac{d^2 y}{dx^2} = ay$,we get $a = 4$.

If $y = 100 e^{2x} + 200 e^{-2x}$ and $\frac{d^2 y}{dx^2} = ay$,then $a = $ . . . . . .

Similar Questions

Let $f(x)$ be a polynomial in $x$. Then the second derivative of $f(e^x)$ is:

Let $f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$,$x \in [0, \pi] - \{\frac{\pi}{4}\}$. Then $f(\frac{7\pi}{12}) f''(\frac{7\pi}{12})$ is equal to

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 $y^2 = a x^2 + 2 x + c$,then $y^3 \frac{d^2 y}{d x^2}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module