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

Answer 1

Cameron Alexander

#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#

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Answer 2

Samantha Adkison

To differentiate ( f(x) = x + \ln(x^2) - x^2 ) using the sum rule, you differentiate each term separately.

For ( x ), the derivative is ( 1 ).
For ( \ln(x^2) ), the derivative is ( \frac{1}{x^2} \cdot 2x = \frac{2}{x} ).
For ( -x^2 ), the derivative is ( -2x ).

So, the derivative of ( f(x) = x + \ln(x^2) - x^2 ) using the sum rule is ( f'(x) = 1 + \frac{2}{x} - 2x ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.