How do you find the derivative for f(x)=cotx/sinx?

Jul 31, 2015

${y}^{'} = - \frac{1 + {\cos}^{2} x}{\sin} ^ 3 x$

Explanation:

You can differentiate this function by using the quotient rule, which tells you that the derivative of afunction expressed as the quotient of two other functions

$f \left(x\right) = g \frac{x}{h} \left(x\right)$

can be found by using

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{{g}^{'} \left(x\right) \cdot h \left(x\right) - g \left(x\right) \cdot {h}^{'} \left(x\right)}{g \left(x\right)} ^ 2}$, with $h \left(x\right) \ne 0$.

You can simplify this expression by using the trigonometric identity

$\cot \left(x\right) = \frac{1}{\tan} \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$

This means that you can write

$f \left(x\right) = \cos \frac{x}{\sin} x \cdot \frac{1}{\sin} x = \cos \frac{x}{\sin} ^ 2 x$

This function's derivative will thus be

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left(\cos x\right)\right] \cdot {\sin}^{2} x - \cos x \cdot \frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right)}{{\sin}^{2} x} ^ 2$

You can use the power and chain rules to find $\frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right)$.

More specifically, you can write ${\sin}^{2} x = {u}^{2}$, with $u = \sin x$. This will get you

$\frac{d}{\mathrm{dx}} \left({u}^{2}\right) = \frac{d}{\mathrm{du}} \left({u}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right)$

$\frac{d}{\mathrm{dx}} \left({u}^{2}\right) = 2 u \cdot \cos x$

$\frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right) = 2 \sin x \cdot \cos x$

Plug this back into your target derivative to get

${f}^{'} = \frac{- \sin x \cdot {\sin}^{2} x - \cos x \cdot 2 \cdot \sin x \cdot \cos x}{\sin} ^ 4 x$

${f}^{'} = \frac{- {\sin}^{3} x - 2 \sin x {\cos}^{2} x}{\sin} ^ 4 x$

You can simplify this further by

${f}^{'} = - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\sin x}}} \left({\sin}^{2} x + 2 {\cos}^{2} x\right)}{\sin} ^ \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \textcolor{b l u e}{3}\right) x$

${f}^{'} = - \frac{{\overbrace{{\sin}^{2} x + {\cos}^{2} x}}^{\textcolor{red}{\text{=1}}} + {\cos}^{2} x}{\sin} ^ 3 x$

${f}^{'} = \textcolor{g r e e n}{- \frac{1 + {\cos}^{2} x}{\sin} ^ 3 x}$