Question

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

Answer 1

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

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