How do you differentiate f(x)= (1-3x^2)^4* (1-x+7x^2)^4  using the product rule?

May 27, 2017

$f ' \left(x\right) = 4 {\left(1 - x + 7 {x}^{2}\right)}^{3} {\left(1 - 3 {x}^{2}\right)}^{3} \left(- 84 {x}^{3} + 9 {x}^{2} + 8 x - 1\right)$

Explanation:

$f \left(x\right) = {\left(1 - 3 {x}^{2}\right)}^{4} \cdot {\left(1 - x + 7 {x}^{2}\right)}^{4}$
Using product rule,
$f ' \left(x\right) = {\left(1 - 3 {x}^{2}\right)}^{4} \cdot \frac{d}{\mathrm{dx}} {\left(1 - x + 7 {x}^{2}\right)}^{4} + {\left(1 - x + 7 {x}^{2}\right)}^{4} \cdot \frac{d}{\mathrm{dx}} {\left(1 - 3 {x}^{2}\right)}^{4}$
Now,
$\frac{d}{\mathrm{dx}} {\left(1 - x + 7 {x}^{2}\right)}^{4} = 4 {\left(1 - x + 7 {x}^{2}\right)}^{3} \cdot \frac{d}{\mathrm{dx}} \left(1 - x + 7 {x}^{2}\right) = 4 {\left(1 - x + 7 {x}^{2}\right)}^{3} \left(14 x - 1\right)$
and,
$\frac{d}{\mathrm{dx}} {\left(1 - 3 {x}^{2}\right)}^{4} = 4 {\left(1 - 3 {x}^{2}\right)}^{3} \cdot \frac{d}{\mathrm{dx}} \left(1 - 3 {x}^{2}\right) = 4 {\left(1 - 3 {x}^{2}\right)}^{3} \left(- 6 x\right)$
Put these two values in the expression for f'(x) and extract the common terms to get,
$f ' \left(x\right) = 4 {\left(1 - x + 7 {x}^{2}\right)}^{3} {\left(1 - 3 {x}^{2}\right)}^{3} \left[\left(1 - 3 {x}^{2}\right) \left(14 x - 1\right) - 6 x \left(1 - x + 7 {x}^{2}\right)\right]$
Further solving the equation we get, $f ' \left(x\right) = 4 {\left(1 - x + 7 {x}^{2}\right)}^{3} {\left(1 - 3 {x}^{2}\right)}^{3} \left(- 84 {x}^{3} + 9 {x}^{2} + 8 x - 1\right)$
which is the final answer