The differential coefficient of sqrt{sec sqrt | vedclass

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

Question

(A) Let $y = \sqrt{\sec \sqrt{x}} = (\sec \sqrt{x})^{1/2}$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{2}(\

Vedclass · Accepted Answer

$\frac{1}{4\sqrt{x}}(\sec \sqrt{x})^{3/2} \sin \sqrt{x}$

Vedclass · Answer

(A) Let $y = \sqrt{\sec \sqrt{x}} = (\sec \sqrt{x})^{1/2}$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{2}(\sec \sqrt{x})^{-1/2} \cdot \frac{d}{dx}(\sec \sqrt{x})$$= \frac{1}{2\sqrt{\sec \sqrt{x}}} \cdot (\sec \sqrt{x} 	an \sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x})$$= \frac{1}{2\sqrt{\sec \sqrt{x}}} \cdot \sec \sqrt{x} 	an \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$$= \frac{1}{4\sqrt{x}} \cdot \frac{\sec \sqrt{x}}{\sqrt{\sec \sqrt{x}}} \cdot 	an \sqrt{x}$$= \frac{1}{4\sqrt{x}} \cdot (\sec \sqrt{x})^{1/2} \cdot \frac{\sin \sqrt{x}}{\cos \sqrt{x}}$$= \frac{1}{4\sqrt{x}} \cdot (\sec \sqrt{x})^{1/2} \cdot \sec \sqrt{x} \cdot \sin \sqrt{x}$$= \frac{1}{4\sqrt{x}} (\sec \sqrt{x})^{3/2} \sin \sqrt{x}$.

The differential coefficient of $\sqrt{\sec \sqrt{x}}$ is

Similar Questions

If $A = \frac{2^x \cot x}{\sqrt{x}}$,then $\frac{dA}{dx} = $

If $y=\log \left(\frac{1+x}{1-x}\right)^{1 / 4}-\frac{1}{2} \tan ^{-1}(x)$,then $\frac{d y}{d x}$ at $x=\frac{1}{\sqrt{2}}$ equals

$\frac{d}{dx}\sqrt{\sec^2 x + \text{cosec}^2 x} = $

If $h(x) = \sqrt{4f(x) + 3g(x)}$,$f(1) = 4$,$g(1) = 3$,$f'(1) = 4$,and $g'(1) = 3$,then find $h'(1)$.

If $y=a \sin x+b \cos x$ (where $a$ and $b$ are constants),then $y^2+\left(\frac{d y}{d x}\right)^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module