Re: Quotient Rule
jml14 said:
y = x^2
3radical (x^2 - 4)
By the way that 3 would look like this if I were to take the derivative it would be (x^2 - 4)^1/3
I keep getting stuck when I have to factor out the negative exponent.
y'(x)= (x^2 - 4)^1/3 (2x) - (x^2) [(1/3 (x^2 - 4) ^ -2/3 (3x^2)]
(x^2 - 4) ^ 7/3
y'(x)= (x^2 - 4)^1/3 (2x) - (x^2) (x^2 - 4) ^ -2/9 (x^2)]
(x^2 - 4) ^ 7/3
It looks like you mean:
\(\displaystyle y =\frac{x^2}{\sqrt[3]{x^2-4}}\)
In that case it looks like you started well, with a couple little errors. I'll try to do this out fully for you, step by step:
You have \(\displaystyle y = \frac{f(x)}{g(x)}\) where:
\(\displaystyle f(x) = x^2\)
\(\displaystyle g(x) = \sqrt[3]{x^2-4} = (x^2-4)^{\frac{1}{3}}\)
The formula for the quotient rule is:
\(\displaystyle y' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\)
We need \(\displaystyle f'(x)\) and \(\displaystyle g'(x)\) and then all we need to do is "plug in."
\(\displaystyle f'(x)=\frac{d}{dx}x^2 = 2x\)
\(\displaystyle g'(x) = \frac{d}{dx}(x^2-4)^{\frac{1}{3}} = \frac{1}{3}(x^2-4)^{\frac{-2}{3}}(2x)\)
Now, using the quotient rule formula we have:
\(\displaystyle y' = \frac{[2x][(x^2-4)^{\frac{1}{3}}]-[x^2][\frac{1}{3}(x^2-4)^{\frac{-2}{3}}(2x)]}{[(x^2-4)^{\frac{1}{3}}]^2}\)
\(\displaystyle y' = \frac{2x(x^2-4)^{\frac{1}{3}}-\frac{2}{3}x^3(x^2-4)^{-\frac{2}{3}}}{(x^2-4)^{\frac{2}{3}}}\)
Now the caluclus is done, we need to simplify. take out the SMALLEST power of each factor you see.
I'll be factoring out an \(\displaystyle x\), a \(\displaystyle \frac{2}{3}\) and \(\displaystyle (x^2-4)^{-\frac{2}{3}}\), all together: \(\displaystyle \frac{2}{3}x(x^2-4)}^{-\frac{2}{3}}\)
So we divide each term in the numerator by this factor, and leave it on the outside:
\(\displaystyle y' = \frac{\frac{2}{3}x(x^2-4)^{-\frac{2}{3}} [3(x^2-4) - x^2]}{(x^2-4)^\frac{2}{3}}\)
The negative exponent dropps its term to the bottom and we add the exponents, and the 3 will also go. The rest of the numerator will also simplify much:
\(\displaystyle y' = \frac{2x[3x^2-12-x^2]}{3(x^2-4)^{\frac{2}{3}+\frac{2}{3}}}\)
\(\displaystyle y' = \frac{2x[2x^2-12]}{3(x^2-4)^{\frac{4}{3}}}\)
\(\displaystyle y' = \frac{4x^3-24x}{3(x^2-4)^{\frac{4}{3}}}\)
Its possible your answer key may look a little different:
\(\displaystyle y' = \frac{4x(x^2-6)}{3(x^2-4)^{\frac{4}{3}}}\)
Hope I've helped :mrgreen:
