Hello again,
I'm again having some trouble using rules to solve down for a derivative of a function. Could someone please show me how I'd find the derivative of
$\displaystyle y=(\frac{4x^2}{3-x})^3$
That would be great. Thanks for reading!
Hello again,
I'm again having some trouble using rules to solve down for a derivative of a function. Could someone please show me how I'd find the derivative of
$\displaystyle y=(\frac{4x^2}{3-x})^3$
That would be great. Thanks for reading!
This is a chain rule problem.
Let's approach it this way. Let $\displaystyle u=\frac{4x^2}{3-x}$. It follows that $\displaystyle y=u^3$. Now by chain rule, $\displaystyle \frac{\,dy}{\,dx}=\frac{\,dy}{\,du}\cdot\frac{\,du }{\,dx}$.
Since we let $\displaystyle u=\frac{4x^2}{3-x}$, it follows by quotient rule that $\displaystyle \frac{\,du}{\,dx}=\frac{(3-x)(8x)-4x^2(-1)}{(3-x)^2}=\frac{24x-4x^2}{(3-x)^2}$.
Therefore,
$\displaystyle \frac{\,dy}{\,dx}=\underbrace{3u^2}_{\frac{\,dy}{\ ,du}}\cdot \underbrace{\frac{24x-4x^2}{(3-x)^2}}_{\frac{\,du}{\,dx}}=3\left(\frac{4x^2}{3-x}\right)^2\cdot\frac{24x-4x^2}{(3-x)^2}$
This can be simplified as $\displaystyle \frac{48x^4\cdot4x\left(6-x\right)}{\left(3-x\right)^4}=\frac{192x^5\left(6-x\right)}{\left(3-x\right)^4}$
Does this make sense?
First note that this is function composition
$\displaystyle f(x)=x^3,g(x)=\frac{4x^2}{3-x}$
$\displaystyle y=f(g(x))=\left(\frac{4x^2}{3-x}\right)^3$
before we use the chain rule lets calculate the needed derviatves
$\displaystyle f'(x)=3x^2$
$\displaystyle g'(x)=\frac{(3-x)(8x)-(4x^2)(-1)}{(3-x)^2}=\frac{24x-8x^2+4x^2}{(3-x)^2}=\frac{24x-4x^2}{(3-x)^2}=\frac{4x(6-x)}{(3-x)^2}$
Now the chain rule tells us that
$\displaystyle y'=f'(g(x))\cdot g'(x)=3\left(\frac{4x^2}{(3-x)} \right)^2\left( \frac{4x(6-x)}{(3-x)^2}\right)=\frac{192x^5(6-x)}{(3-x)^3}$