# How do you differentiate f(x)=1/sqrt(x-3x^3+5x^5 using the quotient rule?

Jan 30, 2017

$f ' \left(x\right) = - \frac{1 - 9 {x}^{2} + 25 {x}^{4}}{2 {\left(x - 3 {x}^{3} + 5 {x}^{5}\right)}^{\frac{3}{2}}} .$

#### Explanation:

Let $f \left(x\right) = \frac{1}{\sqrt{t}} = {t}^{- \frac{1}{2}} , w h e r e , t = x - 3 {x}^{3} + 5 {x}^{5}$.

So, $f \left(x\right) = {t}^{-} \left(\frac{1}{2}\right) , t = x - 3 {x}^{3} + 5 {x}^{5} ,$ meaning that,

$f \text{ is a function of "t, and, t" is of } x$.

In such cases, we apply The Chain Rule, which states that,

$f ' \left(x\right) = \left(\frac{\mathrm{df}}{\mathrm{dt}}\right) \left(\frac{\mathrm{dt}}{\mathrm{dx}}\right) \ldots \ldots \ldots \ldots . . \left(\star\right)$

Knowing that, $\frac{d}{\mathrm{dt}} \left({t}^{n}\right) = n {x}^{n - 1} , \text{ we have,} \frac{\mathrm{df}}{\mathrm{dt}} = - \frac{1}{2} {t}^{- \frac{1}{2} - 1}$ &,

$t = x - 3 {x}^{3} + 5 {x}^{5} \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}} = 1 - 3 \left(3 {x}^{2}\right) + 5 \left(5 {x}^{4}\right) = 1 - 9 {x}^{2} + 25 {x}^{4}$

Altogether,

$f ' \left(x\right) = \left(- \frac{1}{2} {t}^{- \frac{3}{2}}\right) \left(1 - 9 {x}^{2} + 25 {x}^{4}\right) , \mathmr{and} ,$

$f ' \left(x\right) = - \frac{1 - 9 {x}^{2} + 25 {x}^{4}}{2 {\left(x - 3 {x}^{3} + 5 {x}^{5}\right)}^{\frac{3}{2}}}$.