How do you differentiate f(x)=1/ln(sinx) using the quotient rule?

Jul 13, 2016

Step 1: Use the chain rule to differentiate the denominator:

Let $y = \ln \left(u\right)$ and $u = \sin x$.

Then $y ' = \frac{1}{u}$ and $u ' = \cos x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \times \cos x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{x}{\sin} x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cot x$

Now, we have the derivative of the denominator, which will be useful when we use the quotient rule.

Step 2: Use the quotient rule:

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

Then $g \left(x\right) = 1$ and $h \left(x\right) = \ln \left(\sin x\right)$. The derivative of $g \left(x\right)$, since it's a constant is $0$ and the derivative of $h \left(x\right)$ is $\cot x$, as shown in step 1.

The quotient rule states that $f ' \left(x\right) = \frac{g ' \left(x\right) \times h \left(x\right) - g \left(x\right) \times h ' \left(x\right)}{h \left(x\right)} ^ 2$.

$f ' \left(x\right) = \frac{0 \times \ln \left(\sin x\right) - 1 \times \cot x}{\ln \left(\sin x\right)} ^ 2$

$f ' \left(x\right) = - \frac{\cot x}{\ln \left(\sin x\right)} ^ 2$

Hopefully this helps!