Find the derivative

[animal.luver01]

Find the derivative of [x(x-1)^(3/2)]/?(x+1)

So far, I have this:

y'=(x+1)^1/2[x(x(3/2(x-1)^1/2)+(x-1)^3/2]-[x(x-1)^3/2(1/2(x+1)^-1/2)] all over [(x+1)^1/2]^2
y'=(x+1)^1/2[3x/2(x(x-1)^1/2)+(x-1)^3/2]-[1x/2(x-1)^3/2(x+1)^-1/2)] all over (x+1)

That's all I got (which is pretty sad) and I'm supposed to simplify to one term.

 

[galactus]

One way would be to rewrite as a product:

\(\displaystyle x(x-1)^{\frac{3}{2}}(x+1)^{\frac{-1}{2}}\)

Now, use the product rule for 3 terms.

\(\displaystyle f'\cdot g\cdot h+f\cdot g'\cdot h+f\cdot g\cdot h'\)

\(\displaystyle f=x, \;\ g=(x-1)^{\frac{3}{2}}, \;\ h=(x+1)^{\frac{-1}{2}}\)

\(\displaystyle (1)(x-1)^{\frac{3}{2}}(x+1)^{\frac{-1}{2}}+x\left(\frac{3}{2}(x-1)^{\frac{1}{2}}\right)(x+1)^{\frac{-1}{2}}+x(x-1)^{\frac{3}{2}}\left(\frac{-1}{2}(x+1)^{\frac{-3}{2}}\right)\)

\(\displaystyle =\frac{2x^{2}\sqrt{x-1}+2x\sqrt{x-1}-\sqrt{x-1}}{(x+1)^{\frac{3}{2}}}\)

\(\displaystyle =\frac{\sqrt{x-1}\left(2x^{2}+2x-1\right)}{(x+1)^{\frac{3}{2}}}\)