# How do you find the derivative of y=ln(t)/t^2?

Aug 19, 2014

To solve this problem, one must use the quotient rule.

The quotient rule for derivatives states the following:

$\left(\frac{d}{\mathrm{dx}}\right) \left[\frac{u \left(x\right)}{v \left(x\right)}\right] = \frac{u ' \left(x\right) v \left(x\right) - u \left(x\right) v ' \left(x\right)}{{v}^{2} \left(x\right)}$.

In other words, the derivative of the quotient is equal to:

The derivative of the numerator function times the original denominator function, minus the product of the original numerator function and the derivative of the denominator function, all divided by the square of the original denominator function. Note that this only works at locations where $v \left(x\right) \ne 0$

Now, for the particular problem in question, we have $y \left(t\right) = \ln \frac{t}{t} ^ 2$. We can use the steps above, this time using $t$ instead of $x$ as our independent variable. In this case, we have the following:

$u \left(t\right) = \ln \left(t\right) , v \left(t\right) = {t}^{2}$

The derivative of the function $\ln \left(t\right)$ is $\frac{\mathrm{dt}}{t}$, or in this case $\frac{1}{t}$ because our $t$ is simply the variable itself. Further, the derivative of $v \left(t\right)$ can be found using the power rule to be $v ' \left(t\right) = 2 t$. Therefore we also have:

$u ' \left(t\right) = \frac{1}{t} , v ' \left(t\right) = 2 t$

Using our above formula for the quotient rule, we arrive at, for all $t$ such that $v \left(t\right) \ne 0$:

$y ' \left(t\right) = \frac{\left(\frac{1}{t}\right) \cdot \left({t}^{2}\right) - \left(\ln \left(t\right) \cdot 2 t\right)}{{t}^{4}} = \frac{1 - 2 \ln \left(t\right)}{{t}^{3}}$. Note that the simplification of this solution only works for $t \ne 0$, but since our $v \left(t\right) = 0$ at $t = 0$, this is acceptable. Simply include a line stating that this derivative is true for all $x \ne 0$