Question

How do I find the derivative of `ln(e^(4x)+3x)`?

Answer 1

`(f(g(x)))'=(4e^(4x)+3)/(e^(4x)+3x)`

Explanation:

We can find the derivative of this function using chain rule that says:

`color(blue)((f(g(x)))'=f'(g(x))*g'(x))`

Let us decompose the given function into two functions `f(x)` and `g(x)` and find their derivatives as follows:

`g(x)=e^(4x)+3x`
`f(x)=ln(x)`

Let's find the derivative of `g(x)`
Knowing the derivative of exponential that says:
`(e^(u(x)))'=(u(x))'*e^(u(x))`
So,
`(e^(4x))'=(4x)'*e^(4x)=4e^(4x)`
Then ,
`color(blue)(g'(x)=4e^(4x)+3)`

Now Lets find `f'(x)`

`f'(x)=1/x`
According to the property above we have to find `f'(g(x))` so let's substitute `x` by `g(x)` in `f'(x)` we have:

`f'(g(x))=1/g(x)`
`color(blue)(f'(g(x))=1/(e^(4x)+3x))`
Therefore,
`(f(g(x)))'=(1/(e^(4x)+3x))*(4e^(4x)+3)`

`color(blue)((f(g(x)))'=(4e^(4x)+3)/(e^(4x)+3x))`