 $\displaystyle y={{\cot }^{2}}x-5+\cos ecx$

It is difficult to differentiate cot x and cosec x directly. You need to change to a form where you can differentiate  cot x = $\displaystyle \frac{1}{{\tan x}}$ and cosec x = $\displaystyle \frac{1}{{\sin x}}$. $\displaystyle y=\frac{1}{{{{{\tan }}^{2}}x}}-5+\frac{1}{{\sin x}}$ $\displaystyle y={{\tan }^{{-2}}}x-5+{{\sin }^{{-1}}}x$

Use chain rule to differentiate the equation. Don’t forget to differentiate tan x to $\displaystyle {{\sec }^{2}}x$ and sin x to $\displaystyle \cos x$ $\displaystyle \frac{{dy}}{{dx}}=-2{{\tan }^{{-3}}}x\times {{\sec }^{2}}x-{{\sin }^{{-2}}}x\times \cos x$ $\displaystyle \frac{{dy}}{{dx}}=\frac{{-2}}{{{{{\tan }}^{3}}x}}\times \frac{1}{{{{{\cos }}^{2}}x}}-\frac{1}{{{{{\sin }}^{2}}x}}\times \cos x$ $\displaystyle \frac{{dy}}{{dx}}=\frac{{-2}}{{{{{\tan }}^{3}}x{{{\cos }}^{2}}x}}-\frac{{\cos x}}{{{{{\sin }}^{2}}x}}$

Additional Math and Combine Science (Physics/ Chemistry) Tuition at Woodlands, Choa Chu Kang, Yew Tee, Sembawang and Yishun.