If y = cos^2left(frac{5x}{2}right) - sin^2lef | vedclass

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

Question

(A) We know the trigonometric identity $\cos^2 	heta - \sin^2 	heta = \cos(2	heta)$.Given $y = \cos^2\left(\frac{5x}{2}ight) - \sin^2\left(\frac{

Vedclass · Accepted Answer

$-25y$

Vedclass · Answer

(A) We know the trigonometric identity $\cos^2 	heta - \sin^2 	heta = \cos(2	heta)$.Given $y = \cos^2\left(\frac{5x}{2}ight) - \sin^2\left(\frac{5x}{2}ight)$,we can substitute $	heta = \frac{5x}{2}$.Thus,$y = \cos\left(2 	imes \frac{5x}{2}ight) = \cos(5x)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = -\sin(5x) 	imes \frac{d}{dx}(5x) = -5\sin(5x)$.Differentiate again with respect to $x$ to find the second derivative:$\frac{d^2y}{dx^2} = -5 	imes \cos(5x) 	imes \frac{d}{dx}(5x) = -5 	imes 5 \cos(5x) = -25\cos(5x)$.Since $y = \cos(5x)$,we have $\frac{d^2y}{dx^2} = -25y$.

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

Similar Questions

If $y=a \cos 3 x+b e^{-x}$, then $y^{\prime \prime}(3 \sin 3 x-\cos 3 x)=$

If $f: R \rightarrow R$ is an even function which is twice differentiable on $R$ and $f^{\prime \prime}(\pi)=1$, then $f^{\prime \prime}(-\pi)$ is equal to

If $x^2y + y^3 = 2$,then the value of $\frac{d^2y}{dx^2}$ at the point $(1, 1)$ is:

The equation $x^5 - 5x^3 + 5x^2 - 1 = 0$ has how many equal roots?

$\frac{d^n}{dx^n}(e^{2x} + e^{-2x}) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module