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

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Answer 1 · 2017-08-27T13:30:05

$f'(x)=2/x$

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 · 2017-08-29T02:27:00

$f'(x)=2/x$

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 $e^f(x)=x^2$ :

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

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