How do you differentiate e^((x^2-x)^2)  using the chain rule?

Jun 19, 2016

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

Explanation:

Here $f \left(x\right) = {e}^{g \left(x\right)}$, where $g \left(x\right) = {\left(h \left(x\right)\right)}^{2}$ and $h \left(x\right) = {x}^{2} - x$

According to chain rule $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}} \times \frac{\mathrm{dh}}{\mathrm{dx}}$

Hence $\frac{\mathrm{df}}{\mathrm{dx}} = {e}^{{\left({x}^{2} - x\right)}^{2}} \times 2 \left({x}^{2} - x\right) \times \left(2 x - 1\right)$

= ${e}^{{\left({x}^{2} - x\right)}^{2}} \times 2 x \left(x - 1\right) \times \left(2 x - 1\right)$

= $2 x \left(x - 1\right) \left(2 x - 1\right) {e}^{{\left({x}^{2} - x\right)}^{2}}$