# How do you differentiate f(x)= (x^3-1)/( x^2+3x) ^ (3/2)  using the quotient rule?

Oct 8, 2017

${f}^{'} \left(x\right) = \frac{3 {x}^{2} \left({x}^{2} + 3 x\right) - \frac{3}{2} \left({x}^{3} - 1\right) \left(2 x + 3\right)}{{\left({x}^{2} + 3 x\right)}^{\frac{5}{2}}}$

#### Explanation:

$f \left(x\right) = \frac{{x}^{3} - 1}{{x}^{2} + 3 x} ^ \left(\frac{3}{2}\right)$ .Quotient rule: $\frac{d}{\mathrm{dx}} \left(\frac{f}{g}\right) = \frac{g {f}^{'} - f {g}^{'}}{g} ^ 2$

 f^'(x) = (3x^2(x^2+3x)^(3/2)-(x^3-1)*3/2(x^2+3x)^(3/2-1)(2x+3)) /(((x^2+3x)^(3/2))^2

${f}^{'} \left(x\right) = \frac{3 {x}^{2} {\left({x}^{2} + 3 x\right)}^{\frac{3}{2}} - \left({x}^{3} - 1\right) \cdot \frac{3}{2} {\left({x}^{2} + 3 x\right)}^{\frac{1}{2}} \left(2 x + 3\right)}{{\left({x}^{2} + 3 x\right)}^{3}}$

${f}^{'} \left(x\right) = \frac{{\left({x}^{2} + 3 x\right)}^{\frac{1}{2}} \left(3 {x}^{2} \left({x}^{2} + 3 x\right) - \left({x}^{3} - 1\right) \cdot \frac{3}{2} \left(2 x + 3\right)\right)}{{\left({x}^{2} + 3 x\right)}^{3}}$

${f}^{'} \left(x\right) = \frac{3 {x}^{2} \left({x}^{2} + 3 x\right) - \frac{3}{2} \left({x}^{3} - 1\right) \left(2 x + 3\right)}{{\left({x}^{2} + 3 x\right)}^{\frac{5}{2}}}$ [Ans]