Question

Find the derivative of the function.
G(y) = (y − 2)^4/(y^2 + 4y)^9

Answer 1

\[G(y)=\frac{(y-2)^4}{(y^2+4y)^9}=\frac{f(x)}{g(x)^9}=\frac{f(x)}{h(x)}\]
So we will need the chain and quotient rule.
We know that:
\[\frac{d}{dx}(f\circ g)(x)=f'(g(x))g'(x)\phantom{...}(chain\phantom{.}rule)\]
We also know that:
\[\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\]
You remember these?

Answer 2

which one would I use first?

Answer 3

Well we have initially:
\[\frac{d}{dy}G(y)=\frac{d}{dx}\frac{f(y)}{h(y)}=\frac{f'(y)h(y)-h'(y)f(y)}{h(y)^2}\]
Then we can make the substitution: \(h(y)=[g(y)]^9\)
\[\frac{d}{dx}G(y)=\frac{f'(y)[g(y)]^9-[g(y)^9]'f(y)}{[g(y)^9]^2}=\frac{f'(y)[g(y)]^9-[g(y)^9]'f(y)}{g(y)^{18}}\]
Now, it's no issue subbing \(g(y)\) into \(f'(y)[g(y)^9]\) but before we do that into the second term. We can simplify \([g(y)^9]'\) into \(9g(y)^8g'(y)\)
So then i'll simplify a little bit:
\[\frac{f'(y)[g(y)]^9-[g(y)^9]'f(y)}{g(y)^{18}}=\frac{f'(y)(y^2+4y)^9-9(y^2+4y)^8g'(y)(y-2)^4}{(y^2+4y)^{18}}\]

Answer 4

Now all that's left to do is to calculate f'(y) and g'(y) and we're done! Following so far?

Answer 5

Yes, then after that it the chain rule will be used?

Answer 6

Ehh well no! There's no chaining needed to find f'(y) or g'(y)! Where I used the chain rule was in deriving \(g(y)^9\) and obtaining \(\frac{d}{dx}g(y)^9=9g(y)^8g'(y)\)

Answer 7

So we have:
\[f(y)=(y-2)^4\rightarrow f'(y)=4(y-2)^3\]
And
\[g(y)=y^2+4y\rightarrow g'(y)=2y+4\]
So we can substitute both these into the big equation. Therefore:
\[\frac{d}{dx}G(y)=\frac{4(y-2)^3(y^2+4y)^9-9(y^2+4y)^8(2y+4)(y-2)^4}{(y^2+4y)^{18}}\]
And we can simplify a bit:
\[=\frac{4(y-2)^3}{(y^2+4y)^9}-\frac{18(y+2)(y-2)^4}{(y^2+4y)^{10}}\]

Answer 8

And I mean if you REALLY wanted to go ahead, then you would simplify it to this:
\[\eqalign{\frac{d}{dy}G(y)&=(y-2)^3\left[\frac{4}{(y^2+4y)^9}-\frac{18(y+2)(y-2)}{(y^2+4y)^{10}}\right] \\
&=(y-2)^3\left[\frac{4}{(y^2+4y)^9}-\frac{18(y^2-4)}{(y^2+4y)^{10}}\right]
}\]

Answer 9

It's a little nicer :)

Answer 10

AAAND THAT was a messy question haha

Answer 11

Engineering Math for ya!

Answer 12

Haha just you wait till multiple integrals and surface area and volume of 3d figures using integration...THATS MESSY lol 's all part of a good experience