If y=sqrt{2 x+cos ^2left(2 x+frac{pi}{4}right | vedclass

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

Question

(A) Given,$y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)}$.Squaring both sides,we get $y^2=2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)$.Differentiat

Vedclass · Accepted Answer

$\frac{2 \sqrt{2}}{\sqrt{\pi+1}}$

Vedclass · Answer

(A) Given,$y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}ight)}$.Squaring both sides,we get $y^2=2 x+\cos ^2\left(2 x+\frac{\pi}{4}ight)$.Differentiating with respect to $x$:$2 y \frac{d y}{d x} = 2 + 2 \cos \left(2 x+\frac{\pi}{4}ight) \cdot \left(-\sin \left(2 x+\frac{\pi}{4}ight)ight) \cdot 2$.Using the identity $\sin(2	heta) = 2\sin	heta \cos	heta$,we have:$2 y \frac{d y}{d x} = 2 - 2 \sin \left(4 x+\frac{\pi}{2}ight) = 2 - 2 \cos(4x)$.So,$\frac{d y}{d x} = \frac{1 - \cos(4x)}{y}$.At $x = \frac{\pi}{4}$,$y = \sqrt{2(\frac{\pi}{4}) + \cos^2(\frac{\pi}{2} + \frac{\pi}{4})} = \sqrt{\frac{\pi}{2} + \sin^2(\frac{\pi}{4})} = \sqrt{\frac{\pi}{2} + \frac{1}{2}} = \sqrt{\frac{\pi+1}{2}}$.Substituting these values:$\left. \frac{d y}{d x} ight|_{x=\frac{\pi}{4}} = \frac{1 - \cos(\pi)}{\sqrt{\frac{\pi+1}{2}}} = \frac{1 - (-1)}{\frac{\sqrt{\pi+1}}{\sqrt{2}}} = \frac{2 \sqrt{2}}{\sqrt{\pi+1}}$.

If $y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)}$,then find $\frac{d y}{d x}$ at $x=\frac{\pi}{4}$.

Similar Questions

If $f(x) = \frac{x}{1+x}$ and $g(x) = f(f(x))$,then $g^{\prime}(x)$ is equal to

Let $f(x) = \mathop {\text{Lim}}\limits_{h \to 0} \frac{1}{h} \int\limits_x^{x + h} \frac{dt}{t + \sqrt{1 + t^2}}$,then $\mathop {\text{Lim}}\limits_{x \to -\infty} x \cdot f(x)$ is

If $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \cot^{-1}\left(\frac{3-2x}{2+3x}\right)$,then $\frac{dy}{dx}$ is equal to

If $y = \frac{a + b{x^{3/2}}}{{x^{5/4}}}$ and $y' = 0$ at $x = 5$,then the ratio $a:b$ is equal to

Let $f(x + y) = f(x)f(y)$ and $f(x) = 1 + \sin(3x)g(x)$,where $g(x)$ is continuous. Then $f'(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module