# How do you differentiate f(x)=x(1-2x+x^2)^(3/5) using the chain rule?

Dec 24, 2015

Chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$.

#### Explanation:

Renaming $u = 1 - 2 x + {x}^{2}$, we have now $f \left(x\right) = x \cdot {u}^{\frac{3}{5}}$

If we could not derivate directly ${\left(1 - 3 x + {x}^{2}\right)}^{\frac{3}{5}}$, now we can do so using chain rule. Also, we see that this is all about a product (first term is $x$ and the second is ${u}^{\frac{3}{5}}$).

To derivate a product, we must remember the product rule, which states that, for a function $y = f \left(x\right) g \left(x\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

So:

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

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

(df(x))/(dx)=(5(1-2x+x^2)+3x)/(5(1-2x+x^2)^(2/5)