Question

d/dx[ln((x-1)^3(x^2+1)^4)]
I tried but could not get the right answer. I used the chain rule.
The book says the answer is

3/(x-1) + 8x/x^2+1
how do I get rid of the powers I get the same but with powers at the bottom.

Answer 1

so we want to differentiate
\[y=\ln[(x-1)^3](x^2+1)^4 \]?

Answer 2

well first of all you can write this as
\[y=3 \ln(x-1) \cdot (x^2+1)^4 \]

Answer 3

you need the product rule to continue

Answer 4

\[y'=(3\ln(x-1))' \cdot (x^2+4)^4 + 3\ln(x-1) \cdot [(x^2+4)^4]'\]

Answer 5

now continue with chain rule

Answer 6

\[y=\ln \left[ \left( x-1 \right)^3\left( x^2+1 \right)^4 \right]=\ln \left( x-1 \right)^3+\ln \left( x^2+1 \right)^4\]
\[=3\ln \left( x-1 \right)+4\ln \left( x^2+1 \right)\]
\[\frac{ dy }{ dx }=\frac{ 3 }{ x-1 }+4*\frac{ 2x }{ x^2+1 }=?\]

Answer 7

is that ln of that whole thing?

Answer 8

ah maybe it is