# Finding the derivative of ln(1 + x)^2

#### coolbeans33

##### New member
I need to find f'(x) when f(x)= ln(1+x)2.

I started with the chain rule:

d[ln(1+x)2]/d(1+x)2 * d(1+x)2/d(1+x) * d(1+x)/dx

so

1/(1+x) * 1/(1+x)2​ * 2(1+x)

I know something about that is wrong, I'm not sure what.

#### MarkFL

Staff member
Re: finding the derivative of ln(1+x)^2

I would first use the log property:

$$\displaystyle \log_a\left(b^c \right)=c\log_a(b)$$

to simplify the function prior to differentiating.

Next, I would use the logarithmic and chain rules:

$$\displaystyle \frac{d}{dx}\left(\ln(u(x)) \right)=\frac{u'(x)}{u(x)}$$

What do you find?

#### coolbeans33

##### New member
Re: finding the derivative of ln(1+x)^2

then it would be f(x)= 2ln(1+x)

and f'(x)=2/(1+x)

So if I want to differentiate f(x)=ln(1+x2)2

it would be 4x/(1+x)2 right?

#### MarkFL

Staff member
Re: finding the derivative of ln(1+x)^2

then it would be f(x)= 2ln(1+x)

and f'(x)=2/(1+x)
Correct.

So if I want to differentiate f(x)=ln(1+x2)2

it would be 4x/(1+x)2 right?
Not quite...it would be:

$$\displaystyle f'(x)=2\left(\frac{1}{1+x^2}\cdot2x \right)=\frac{4x}{1+x^2}$$