Question

Differentiate : Csqrt%7B2x%2B7%7D%7D" id="TexFormula1" title="\mathsf {y = cos^{-1}\frac{x}{2} \div \sqrt{2x+7}}" alt="\mathsf {y = cos^{-1}\frac{x}{2} \div \sqrt{2x+7}}" align="absmiddle" class="latex-formula">

Answer 1

Answer:

$\dfrac{dy}{dx}=-\dfrac{2x+7+\arccos\left(\dfrac{x}{2}\right)\sqrt{4-x^2}}{(2x+7)\sqrt{4-x^2}\sqrt{2x+7}}$

Step-by-step explanation:

Quotient Rule

$\textsf{If }\:\:y=\dfrac{u}{v} \:\textsf{ then}:$

$\implies \dfrac{dy}{dx}=\dfrac{v \dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}$

Given equation:

$y=\dfrac{\arccos\left(\dfrac{x}{2}\right)}{\sqrt{2x+7}}$

To differentiate the given equation, apply the Quotient Rule:

$\textsf{Let }\:u=\arccos\left(\dfrac{x}{2}\right)$

$\textsf{If }y=\arccos x \implies \dfrac{dy}{dx}=-\dfrac{1}{\sqrt{1-x^2}}$

$\begin{aligned}\implies u=\arccos \left(\dfrac{x}{2}\right) \implies \dfrac{du}{dx}& =-\dfrac{1}{\sqrt{1- \left(\dfrac{x}{2}\right)^2}}\dfrac{d}{dx}\left(\dfrac{x}{2}\right)\\\\& =-\dfrac{1}{\sqrt{1- \left(\dfrac{x}{2}\right)^2}} \cdot \dfrac{1}{2}\\\\ & = -\dfrac{1}{2\sqrt{1- \left(\dfrac{x}{2}\right)^2}}\\\\& =-\dfrac{1}{2\sqrt{\dfrac{4-x^2}{4}}}\\\\& =-\dfrac{1}{\dfrac{2\sqrt{4-x^2}}{2}} \\\\ & = -\dfrac{1}{\sqrt{4-x^2}}\end{aligned}$

$\textsf{Let }\:v=\sqrt{2x+7}=(2x+7)^{\frac{1}{2}}$

$\begin{aligned}\implies \dfrac{dv}{dx} & =\dfrac{1}{2}(2x+7)^{-\frac{1}{2}} \cdot 2\\\\ & = (2x+7)^{-\frac{1}{2}}\\\\ & = \dfrac{1}{\sqrt{2x+7}}\end{aligned}$

Therefore:

$\implies \dfrac{dy}{dx}=\dfrac{-\dfrac{\sqrt{2x+7}}{\sqrt{4-x^2}}-\dfrac{\arccos\left(\dfrac{x}{2}\right)}{\sqrt{2x+7}}}{(\sqrt{2x+7})^2}$

$\implies \dfrac{dy}{dx}=\dfrac{-\dfrac{\sqrt{2x+7}}{\sqrt{4-x^2}}-\dfrac{\arccos\left(\dfrac{x}{2}\right)}{\sqrt{2x+7}}}{2x+7}$

$\implies \dfrac{dy}{dx}=\dfrac{-\dfrac{\sqrt{2x+7}\sqrt{2x+7}}{\sqrt{4-x^2}\sqrt{2x+7}}-\dfrac{\arccos\left(\dfrac{x}{2}\right)\sqrt{4-x^2}}{\sqrt{4-x^2}\sqrt{2x+7}}}{2x+7}$

$\implies \dfrac{dy}{dx}=\dfrac{-\dfrac{(2x+7)+\arccos\left(\dfrac{x}{2}\right)\sqrt{4-x^2}}{\sqrt{4-x^2}\sqrt{2x+7}}}{2x+7}$

$\implies \dfrac{dy}{dx}=-\dfrac{(2x+7)+\arccos\left(\dfrac{x}{2}\right)\sqrt{4-x^2}}{\sqrt{4-x^2}\sqrt{2x+7}}}\times\dfrac{1}{2x+7}$

$\implies \dfrac{dy}{dx}=-\dfrac{2x+7+\arccos\left(\dfrac{x}{2}\right)\sqrt{4-x^2}}{(2x+7)\sqrt{4-x^2}\sqrt{2x+7}}$

Answer 2

Refer to the attachment for calculation

$\underline{\boxed{\bf \dfrac{dy}{dx}=\dfrac{-1}{\sqrt{-2x^3-7x^2+8x+28}}-\dfrac{cos^{-1}\dfrac{x}{2}}{\sqrt{(2x+7)^3}}}}$