# How do you find the derivative of f(x) = [(3x^2 + 1)^(1/3) - 5]^2 / [5x^2 + 4]^(1/2)?

Aug 24, 2016

$f ' \left(x\right) = \frac{\frac{4 x {\left(5 {x}^{2} + 4\right)}^{\frac{1}{2}}}{{\left(3 {x}^{2} + 1\right)}^{\frac{2}{3}}} \left\{{\left(3 {x}^{2} + 1\right)}^{\frac{1}{3}} - 5\right\} - \frac{5 x}{5 {x}^{2} + 4} ^ \left(\frac{1}{2}\right) {\left\{{\left(3 {x}^{2} + 1\right)}^{\frac{1}{3}} - 5\right\}}^{2}}{5 {x}^{2} + 4}$

#### Explanation:

We will use the Quotient Rule and Chain Rule.

$f \left(x\right) = {\left\{{\left(3 {x}^{2} + 1\right)}^{\frac{1}{3}} - 5\right\}}^{2} / {\left(5 {x}^{2} + 4\right)}^{\frac{1}{2}} = \frac{u}{v} , s a y ,$, where,

u={(3x^2+1)^(1/3)-5}^2, &, v=(5x^2+4)^(1/2).

Now, $f = \frac{u}{v} \Rightarrow f ' = \frac{v u ' - u v '}{v} ^ 2. \ldots \ldots \ldots \ldots . . \left(\star\right)$.

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

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

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

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

$= \frac{2}{3} \cdot 6 x \left\{{\left(3 {x}^{2} + 1\right)}^{\frac{1}{3}} - 5\right\} {\left(3 {x}^{2} + 1\right)}^{- \frac{2}{3}}$

$u ' = \frac{4 x}{3 {x}^{2} + 1} ^ \left(\frac{2}{3}\right) \left\{{\left(3 {x}^{2} + 1\right)}^{\frac{1}{3}} - 5\right\}$

$v = {\left(5 {x}^{2} + 4\right)}^{\frac{1}{2}}$.

$\Rightarrow v ' = \frac{1}{2} {\left(5 {x}^{2} + 4\right)}^{- \frac{1}{2}} \left(5 {x}^{2} + 4\right) '$

$= \frac{1}{2} \cdot 10 x {\left(5 {x}^{2} + 4\right)}^{- \frac{1}{2}}$

$v ' = \frac{5 x}{5 {x}^{2} + 4} ^ \left(\frac{1}{2}\right)$

Using all these, together with, ${v}^{2} = \left(5 {x}^{2} + 4\right)$, in $\left(\star\right)$, we get,

$f ' \left(x\right) = \frac{\frac{4 x {\left(5 {x}^{2} + 4\right)}^{\frac{1}{2}}}{{\left(3 {x}^{2} + 1\right)}^{\frac{2}{3}}} \left\{{\left(3 {x}^{2} + 1\right)}^{\frac{1}{3}} - 5\right\} - \frac{5 x}{5 {x}^{2} + 4} ^ \left(\frac{1}{2}\right) {\left\{{\left(3 {x}^{2} + 1\right)}^{\frac{1}{3}} - 5\right\}}^{2}}{5 {x}^{2} + 4}$

Enjoy maths.!