How do you differentiate g(x) =(x^2 + 6) (x^2 + 1)^5  using the product rule?

May 11, 2016

$\frac{d}{\mathrm{dx}} g \left(x\right) = 2 x {\left({x}^{2} + 1\right)}^{4} \left(6 {x}^{2} + 31\right)$

Explanation:

$\frac{d}{\mathrm{dx}} \left(p \left(x\right) \cdot q \left(x\right)\right) = \frac{\mathrm{dp}}{\mathrm{dx}} \cdot q \left(x\right) + \frac{\mathrm{dq}}{\mathrm{dx}} \cdot p \left(x\right)$

Hence $\frac{d}{\mathrm{dx}} g \left(x\right) = \left(2 x\right) \cdot {\left({x}^{2} + 1\right)}^{5} + \left({x}^{2} + 6\right) \cdot 5 {\left({x}^{2} + 1\right)}^{4} \cdot 2 x$

= $2 x {\left({x}^{2} + 1\right)}^{5} + 10 x \left({x}^{2} + 6\right) {\left({x}^{2} + 1\right)}^{4}$

= $2 x {\left({x}^{2} + 1\right)}^{4} \left\{\left({x}^{2} + 1\right) + 5 \left({x}^{2} + 6\right)\right\}$

= $2 x {\left({x}^{2} + 1\right)}^{4} \left(6 {x}^{2} + 31\right)$