I need help to find the derivative of $(9x^6+4x^3)^4 $
I already tried the chain rule but I got the wrong answer and I do not know what I did wrong.
This is what I got:
$$4(9x^6+4x^3)^3 \times 54x^5+12x^2$$
I need help to find the derivative of $(9x^6+4x^3)^4 $
I already tried the chain rule but I got the wrong answer and I do not know what I did wrong.
This is what I got:
$$4(9x^6+4x^3)^3 \times 54x^5+12x^2$$
$$f(g)=g^4,\\ g(x)=9x^6+4x^3.$$
Then
$$\frac{df(g)}{dg}=4g^3,\\\frac{dg(x)}{dx}=54x^5+12x^2$$ and
$$\frac{df(g(x))}{dx}=4g^3(x)(6x^5+12x^2).$$
using the chain and the power rule we get $$4(9x^6+4x^3)^4(54x^5+12x^2)$$ simplifying we obtain $$24x^{11}(9x^3+2)(9x^3+4)^3$$
Your answer is merely missing a pair of parentheses around the $54x^5+12x^2$. If you weren't already aware how important parentheses can be, hopefully you are now.
I'm guessing this was an exercise done at a computer, and the computer only told you you got it wrong, without any further explanation. I can't imagine a human grader with a red pen not inking in the correction, even if they are hardnosed and give no partial credit for the attempt.
Apply the chain rule: $\frac{df(y)}{dx}=\frac{df}{dy}\frac{dy}{dx}$.
Let $y=(9x^6+4x^4)$. So you get $\frac{d}{dy}(y^4)=4y^3$ and $\frac{d}{dx}(9x^6+4x^4)=54x^5+12x^2$.
So, $\frac{df}{dy}\frac{dy}{dx}=4y^3(54x^5+12x^2)$. Substitute $y$ back with $(9x^6+4x^4)$ and you get:
$4(9x^6+4x^4)^3(54x^5+12x^2)$