How do you differentiate $f(x)=x+lnx^2-x^2$ using the sum rule?

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Answer 1 · 2015-12-20T17:43:14

$f'(x)=1+2/x-2x$

Rewrite using logarithm rules.

$f(x)=x+2lnx-x^2$

The sum rule simply means that you can take the derivative of each individual part and then add them together.

$d/dx[x]=1$

$d/dx[2lnx]=2d/dx[lnx]=2(1/x)=2/x$

$d/dx[x^2]=2x$

Thus,

$f'(x)=1+2/x-2x$

Simplified:

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

How do you differentiate f(x)=x+lnx^2-x^2f(x)=x+lnx^2-x^2 using the sum rule?