Please help me use the product rule to differentiate?

Elena

New member
Joined
Mar 26, 2013
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1
I need to find the derivative of: f(x)=x[ln(x2+1)]3

From what I understand about the product rule it is: f(x)*g'(x)+f'(x)*g(x)

So I do: x*(3(ln(x2+1)2)*(2x/(x2+1))+(ln(x2+1))3)

and I get:
[(3x(ln(x2+1))2)*2x]/(x2+1)
+ln(x2+1)3

=ln(x2+1)3 +[(6x2(ln(x2+1))2]/[(x2+1)]
---------------------------------------------------------------

But the answer I am supposed to end up with is:

[ln(x2+1)]2
*[ln(x2+1)+((6x2/((x2+1))]

So what did I do wrong ?
icon9.png


Thank you!
 
Your answer is correct.

\(\displaystyle \displaystyle \left[\ln(x^2+1)\right]^3+\frac{6x^2\left[\ln(x^2+1)\right]^2}{x^2+1}\,=\,\left[\ln(x^2+1)\right]^2\left[\ln(x^2+1)+\frac{6x^2}{x^2+1}\right]\)
 
I need to find the derivative of: f(x) = x [ln(x2+1)]3

From what I understand about the product rule it is: f(x)*g'(x) + f'(x)*g(x)

So I do: x*(3(ln(x2+1)2)*(2x/(x2+1)) + (ln(x2+1))3) closing ) in wrong place
........x [3{ln(x²+1)}² {2x/(x²+1)}] + {ln(x²+1)

and I get:
[(3x(ln(x2+1))2)*2x]/(x2+1)
+ln(x2+1)3

=ln(x2+1)3 +[(6x2(ln(x2+1))2]/[(x2+1)] OK . . .
---------------------------------------------------------------

But the answer I am supposed to end up with is:

[ln(x2+1)]2
*[ln(x2+1)+((6x2/((x2+1))]

So what did I do wrong ?
icon9.png


Thank you!
Thank you for showing your work!

Doing great - all you have left to do is factor out the common factor {ln(x²+1)}² from the two terms.
 
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