Find the value of k if frac{d}{d x}left{frac{ | vedclass

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

Question

(A) Let $y = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 x}}}}$.Using the identity $1+\cos 2A = 2 \cos^2 A$,we simplify the expression step by step:$y =

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let $y = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 x}}}}$.Using the identity $1+\cos 2A = 2 \cos^2 A$,we simplify the expression step by step:$y = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 4 x)}}}} = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{4 \cos^2 2 x}}}} = \frac{2}{\sqrt{2+\sqrt{2+2 \cos 2 x}}}$.Continuing the simplification:$y = \frac{2}{\sqrt{2+\sqrt{2(1+\cos 2 x)}}} = \frac{2}{\sqrt{2+\sqrt{4 \cos^2 x}}} = \frac{2}{\sqrt{2+2 \cos x}} = \frac{2}{\sqrt{2(1+\cos x)}} = \frac{2}{\sqrt{4 \cos^2 \frac{x}{2}}}$.Thus,$y = \frac{2}{2 \cos \frac{x}{2}} = \sec \frac{x}{2}$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \sec \frac{x}{2} 	an \frac{x}{2} \cdot \frac{1}{2}$.Comparing this with $k \sec \frac{x}{2} 	an \frac{x}{2}$,we get $k = \frac{1}{2}$.

Find the value of $k$ if $\frac{d}{d x}\left\{\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos (4 x)}}}}\right\} = k \sec \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)$

Similar Questions

Find the derivative of $f(x) = 3$ at $x = 0$ and at $x = 3$.

If $f(0)=0$ and $f^{\prime}(0)=3$,then the derivative of $y=f(f(f(f(f(x)))))$ at $x=0$ is

The values of $x$,at which the first derivative of the function $(\sqrt{x} + \frac{1}{\sqrt{x}})^2$ with respect to $x$ is $\frac{3}{4}$,are

Let $f$ be any function defined on $R$ and let it satisfy the condition $|f(x) - f(y)| \leq |(x - y)^2|$,for all $(x, y) \in R$. If $f(0) = 1$,then:

If $y = (1 + x^{1/4})(1 + x^{1/2})(1 - x^{1/4})$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module