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

[coolbeans33]

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]

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]

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]

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:

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