# How solve it? Derivate This Please!!!!

Mar 17, 2017

$\frac{\mathrm{dF}}{\mathrm{dx}} = \frac{3 {\left(10 {x}^{3} - 7 {x}^{2}\right)}^{2} \left(60 {x}^{3} - 321 {x}^{2} + 140 x\right)}{3 x - 10} ^ 4$

#### Explanation:

As $F \left(x\right) = {\left(\frac{10 {x}^{3} - 7 {x}^{2}}{3 x - 10}\right)}^{3}$, let $G \left(x\right) = \frac{10 {x}^{3} - 7 {x}^{2}}{3 x - 10}$

then we can use chain rule as $F \left(x\right) = {\left[G \left(x\right)\right]}^{3}$.

But before that let us workout $\frac{\mathrm{dG}}{\mathrm{dx}}$

using Quotient rule, which states if $f \left(x\right) = \frac{u \left(x\right)}{v \left(x\right)}$

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{du}}{\mathrm{dx}} \times v \left(x\right) - \frac{\mathrm{dv}}{\mathrm{dx}} \times u \left(x\right)}{v \left(x\right)} ^ 2$

In $G \left(x\right) = \frac{10 {x}^{3} - 7 {x}^{2}}{3 x - 10}$, we have $u \left(x\right) = 10 {x}^{3} - 7 {x}^{2}$ and $v \left(x\right) = 3 x - 10$.

and $\frac{\mathrm{du}}{\mathrm{dx}} = 30 {x}^{2} - 14 x$ and $\frac{\mathrm{dv}}{\mathrm{dx}} = 3$

Hence, $\frac{\mathrm{dG}}{\mathrm{dx}} = \frac{\left(30 {x}^{2} - 14 x\right) \times \left(3 x - 10\right) - 3 \times \left(10 {x}^{3} - 7 {x}^{2}\right)}{3 x - 10} ^ 2$

= $\frac{\left(90 {x}^{3} - 300 {x}^{2} - 42 {x}^{2} + 140 x\right) - \left(30 {x}^{3} - 21 {x}^{2}\right)}{3 x - 10} ^ 2$

= $\frac{60 {x}^{3} - 321 {x}^{2} + 140 x}{3 x - 10} ^ 2$

Now $\frac{\mathrm{dF}}{\mathrm{dx}} = 3 {\left(G \left(x\right)\right)}^{2} \times \frac{\mathrm{dG}}{\mathrm{dx}}$

= $3 {\left(\frac{10 {x}^{3} - 7 {x}^{2}}{3 x - 10}\right)}^{2} \times \frac{60 {x}^{3} - 321 {x}^{2} + 140 x}{3 x - 10} ^ 2$

= $\frac{3 {\left(10 {x}^{3} - 7 {x}^{2}\right)}^{2} \left(60 {x}^{3} - 321 {x}^{2} + 140 x\right)}{3 x - 10} ^ 4$