Question

Differentiate :

`\mathsf {y = cos^(-1)(x)/(2) / √(2x+7)}`

Answer 1

Refer to the attachment for calculation

`\underline{\boxed{\bf (dy)/(dx)=(-1)/(√(-2x^3-7x^2+8x+28))-(cos^(-1)(x)/(2))/(√((2x+7)^3))}}`

Answer 2

`(dy)/(dx)=-(2x+7+\arccos\left((x)/(2)\right)√(4-x^2))/((2x+7)√(4-x^2)√(2x+7))`

Explanation:

Quotient Rule

`\textsf{If }\:\:y=(u)/(v) \:\textsf{ then}:`

`\implies (dy)/(dx)=(v (du)/(dx)-u(dv)/(dx))/(v^2)`

Given equation:

`y=(\arccos\left((x)/(2)\right))/(√(2x+7))`

To differentiate the given equation, apply the Quotient Rule:

`\textsf{Let }\:u=\arccos\left((x)/(2)\right)`

`\textsf{If }y=\arccos x \implies (dy)/(dx)=-(1)/(√(1-x^2))`

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

`\textsf{Let }\:v=√(2x+7)=(2x+7)^{(1)/(2)}`

`\begin{aligned}\implies (dv)/(dx) & =(1)/(2)(2x+7)^{-(1)/(2)} \cdot 2\\\\ & = (2x+7)^{-(1)/(2)}\\\\ & = (1)/(√(2x+7))\end{aligned}`

Therefore:

`\implies (dy)/(dx)=(-(√(2x+7))/(√(4-x^2))-(\arccos\left((x)/(2)\right))/(√(2x+7)))/((√(2x+7))^2)`

`\implies (dy)/(dx)=(-(√(2x+7))/(√(4-x^2))-(\arccos\left((x)/(2)\right))/(√(2x+7)))/(2x+7)`

`\implies (dy)/(dx)=(-(√(2x+7)√(2x+7))/(√(4-x^2)√(2x+7))-(\arccos\left((x)/(2)\right)√(4-x^2))/(√(4-x^2)√(2x+7)))/(2x+7)`

`\implies (dy)/(dx)=(-((2x+7)+\arccos\left((x)/(2)\right)√(4-x^2))/(√(4-x^2)√(2x+7)))/(2x+7)`

`\implies (dy)/(dx)=-((2x+7)+\arccos\left((x)/(2)\right)√(4-x^2))/(√(4-x^2)√(2x+7))}*(1)/(2x+7)`

`\implies (dy)/(dx)=-(2x+7+\arccos\left((x)/(2)\right)√(4-x^2))/((2x+7)√(4-x^2)√(2x+7))`