# How do you differentiate f(t)= sqrt(ln(t)) / (-4t-6)^2 using the quotient rule?

Jan 11, 2018

$f ' \left(t\right) = \frac{\left(- 4 t - 6\right) + 16 t \ln t}{2 t \sqrt{\ln t} {\left(- 4 t - 6\right)}^{3}}$

#### Explanation:

For the quotient rule:

$f \left(t\right) = u v \to f ' \left(t\right) = \frac{u ' v - u v '}{v} ^ 2$

So

u = sqrt(ln(t))-> u'=1/(2tsqrt(ln(t))

And:

$v = {\left(- 4 t - 6\right)}^{2} \to v ' = - 8 \left(- 4 t - 6\right)$

Using the quotient rule:

$f ' \left(t\right) = \frac{\frac{1}{2 t \sqrt{\ln t}} {\left(- 4 t - 6\right)}^{2} + 8 \sqrt{\ln \left(t\right)} \left(- 4 t - 6\right)}{- 4 t - 6} ^ 4$

$= \frac{\frac{{\left(- 4 t - 6\right)}^{2} + 16 t \ln t \left(- 4 t - 6\right)}{2 t \sqrt{\ln t}}}{- 4 t - 6} ^ 4$

Simplifying by canceling a factor of $\left(- 4 t - 6\right)$:

$= \frac{\left(- 4 t - 6\right) + 16 t \ln t}{2 t \sqrt{\ln t} {\left(- 4 t - 6\right)}^{3}}$