# How do you find the derivative of y = 9 ln(4 ln x)?

Dec 26, 2017

Apply the chain rule.

#### Explanation:

Begin with the derivative of the outermost function, $\ln$. We know that the derivative of the natural log is just one over the argument of the function. In this case, the argument is $4 \ln \left(x\right)$, so we begin with $\frac{1}{4 \ln \left(x\right)}$. Of course $9$ is a constant, so we leave it alone.

$\implies y ' = 9 \cdot \frac{1}{4 \ln \left(x\right)} \cdot \text{...}$

Now we multiply by the derivative of the next function, which is $4 \ln \left(x\right)$. Again, we know that this derivative is just $4 \cdot \frac{1}{x}$

$\implies y ' = 9 \cdot \frac{1}{4 \ln \left(x\right)} \cdot 4 \cdot \frac{1}{x} \cdot \text{...}$

Finally, the innermost term is just $x$, which has derivative 1.

$\implies y ' = 9 \cdot \frac{1}{4 \ln \left(x\right)} \cdot 4 \cdot \frac{1}{x} \cdot 1$

$\therefore y ' = \frac{9}{x \ln \left(x\right)}$