Differentiate $y= x(4-9x^4)^4$ So the textbook tells me to do the following question but I can't seem to get the right answer. (answer: $(4-9x^4)^3(4-45x^4)$
This is what I've done: 
apply the product rule
$u = x$
$u' = 1$
$v = (4-9x^4)^4$
$v' = -144x^3 (4-9x^4)^3$
I used the product rule and got: $(4-9x^4)^3 (-153x^4+ 4)$
What am I doing wrong? 
edit: the textbook has the wrong answer, thanks anyways :D 

$y = x\left(4-9x^4\right)^4$
Use the product rule, $\left(uv\right)' = u'v + uv'$, setting:
$u = x$, $v = \left(4-9x^4\right)^4$.
Then,
$u' = 1$
Using the chain rule on $v$, we get:
$v' = 4\left(4-9x^4\right)^3 \cdot \left(-36x^3\right) = -144x^3\left(4-9x^4\right)^3$
Now, $\left(uv\right)' = u'v + uv' = \left(4-9x^4\right)^4 + -144x^4\left(4-9x^4\right)^3 = \left(4-9x^4\right)^3\left(4-9x^4-144x^4\right)$.
The answer is therefore $\left(4-9x^4\right)^3\left(4-153x^4\right)$, so you were correct, assuming the problem was transcribed correctly.
