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

Apr 4, 2016

$6 x \sqrt{2 {x}^{2} + 1} {\sqrt{x - 1}}^{7} + \frac{7}{4} {\sqrt{x - 1}}^{3} {\sqrt{2 {x}^{2} + 1}}^{3}$

Explanation:

The product rule states that if

$g \left(x\right) = s \left(x\right) \cdot t \left(x\right)$

then

$g ' \left(x\right) = s ' \left(x\right) t \left(x\right) + t ' \left(x\right) s \left(x\right)$.

If we substitute $s \left(x\right) = {\left(2 {x}^{2} + 1\right)}^{\frac{3}{2}}$ and $t \left(x\right) = {\left(x - 1\right)}^{\frac{7}{4}}$, we can find $s ' \left(x\right)$ and $t ' \left(x\right)$ using the chain rule, because each function is one function (in the brackets) inside another (the index or power).

The chain rule says that if

$h \left(x\right) = k \left(l \left(x\right)\right) \to h ' \left(x\right) = l ' \left(x\right) k ' \left(l\right)$

Therefore,

$\frac{d}{\mathrm{dx}} \left(2 {x}^{2} + 1\right) = 4 x$
$\frac{d}{\mathrm{dx}} {u}^{\frac{3}{2}} = \frac{3}{2} {u}^{\frac{1}{2}}$, where $u$ can be $\left(2 {x}^{2} + 1\right)$.
$s ' \left(x\right) = 4 x \cdot \frac{3}{2} {\left(2 {x}^{2} + 1\right)}^{\frac{1}{2}} = 6 x \sqrt{2 {x}^{2} + 1}$

and for $t \left(x\right)$,

$\frac{d}{\mathrm{dx}} \left(x - 1\right) = 1$
$\frac{d}{\mathrm{dx}} {u}^{\frac{7}{4}} = \frac{7}{4} {u}^{\frac{3}{4}}$
$t ' \left(x\right) = \frac{7}{4} {\left(x - 1\right)}^{\frac{3}{4}} = \frac{7}{4} {\sqrt{x - 1}}^{3}$

Therefore, by substituting into the product rule,

$s ' \left(x\right) t \left(x\right) + t ' \left(x\right) s \left(x\right) =$
$6 x \sqrt{2 {x}^{2} + 1} {\sqrt{x - 1}}^{7} + \frac{7}{4} {\sqrt{x - 1}}^{3} {\sqrt{2 {x}^{2} + 1}}^{3}$