Question

How do you find `f'(x)` if `f(x) = 2ln(x)`?

Answer 1

`f'(x)=2/x`

Explanation:

We know that the derivative of `lnx` is `1/x`
We have `f (x)=2lnx`
Differentiating both sides with respect to `x` we get
`f'(x)=2/x`

Answer 2

`f'(x)=2/x`

Explanation:

In case you don't know the derivative of `ln(x)`, we can first rewrite the logarithm function using `blog(a)=log(a^b)`:

`f(x)=2ln(x)=ln(x^2)`

Exponentiate both sides with `e` as the base (that is, undo the logarithm):

`e^f(x)=x^2`

Now we can differentiate. Use the chain rule on the left-hand side:

`e^f(x)*f'(x)=2x`

Solve for the derivative:

`f'(x)=(2x)/e^f(x)`

Recall that `e^f(x)=x^2`:

`f'(x)=(2x)/x^2=2/x`