How do you differentiate f(x)=x^7sqrt(4x^2+7) using the product rule?

Sep 1, 2016

$f ' \left(x\right) = 7 {x}^{6} {\left(4 {x}^{2} + 7\right)}^{\frac{1}{2}} + \frac{4 {x}^{8}}{4 {x}^{2} + 7} ^ \left(\frac{1}{2}\right)$

Explanation:

Let $f \left(x\right) = g \left(x\right) \times h \left(x\right)$

The derivative of $f \left(x\right)$, by the product rule, is given by $f ' \left(x\right) = g ' \left(x\right) \times h \left(x\right) + g \left(x\right) \times h ' \left(x\right)$

We must therefore differentiate both $g \left(x\right) \mathmr{and} h \left(x\right)$.

$g ' \left(x\right) = 7 \times {x}^{7 - 1}$

$g ' \left(x\right) = 7 {x}^{6}$

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We differentiate $h \left(x\right)$ using the chain rule.

Letting $y = {u}^{\frac{1}{2}}$ and $u = 4 {x}^{2} + 7$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {u}^{- \frac{1}{2}} \times 8 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 x}{2 {\left(4 {x}^{2} + 7\right)}^{\frac{1}{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 x}{4 {x}^{2} + 7} ^ \left(\frac{1}{2}\right)$

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We now have all the information we need to apply the product rule.

$f ' \left(x\right) = 7 {x}^{6} \times {\left(4 {x}^{2} + 7\right)}^{\frac{1}{2}} + \frac{4 x}{4 {x}^{2} + 7} ^ \left(\frac{1}{2}\right) \times {x}^{7}$

$f ' \left(x\right) = 7 {x}^{6} {\left(4 {x}^{2} + 7\right)}^{\frac{1}{2}} + \frac{4 {x}^{8}}{4 {x}^{2} + 7} ^ \left(\frac{1}{2}\right)$

This can be simplified further, but I'll leave the algebra to you.

Hopefully this helps!