Question

Find f '(x) for f(x) = squaroot cos (6x) Find dy/dx of y = csc squareroot x

Answer 1

$\bf f(x)=\sqrt{cos(6x)}\implies f(x)=[cos(6x)]^{\frac{1}{2}} \\\\\\ f'(x)=\stackrel{\textit{using the chain-rule}}{\cfrac{1}{2}[cos(6x)]^{-\frac{1}{2}}~\cdot ~-sin(6x)~\cdot ~6}\implies f'(x)=\cfrac{-6sin(6x)}{2[cos(6x)]^{\frac{1}{2}}} \\\\\\ f'(x)=\cfrac{-3sin(6x)}{\sqrt{cos(6x)}} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf y=csc(\sqrt{x})\implies y=csc\left( x^{\frac{1}{2}} \right) \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{using the chain-rule}}{-csc\left( x^{\frac{1}{2}} \right)cot\left( x^{\frac{1}{2}} \right)~\cdot ~\cfrac{1}{2}x^{-\frac{1}{2}}}\implies \cfrac{dy}{dx}=-\cfrac{csc\left( x^{\frac{1}{2}} \right)cot\left( x^{\frac{1}{2}} \right)}{2x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=-\cfrac{csc(\sqrt{x})cot(\sqrt{x})}{2\sqrt{x}}$