The differential coefficient of sqrt{sec sqrt | vedclass

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(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}(\

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

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}$.

$I$. $f'(x) = \frac{3}{4}x^{4/3} + 9$	$II$. $f(x) = \frac{9}{28}x^{7/3} - 2$
$III$. $f(x) = \frac{9}{28}x^{7/3} + 6$	$IV$. $f'(x) = \frac{3}{4}x^{4/3} - 4$

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

Similar Questions

Evaluate: $\frac{d}{d x}\left[e^{\tan ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]$

Which of the following statements could be true if $f''(x) = x^{1/3}$?

$I$. $f'(x) = \frac{3}{4}x^{4/3} + 9$ $II$. $f(x) = \frac{9}{28}x^{7/3} - 2$

$III$. $f(x) = \frac{9}{28}x^{7/3} + 6$ $IV$. $f'(x) = \frac{3}{4}x^{4/3} - 4$

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