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

Sep 26, 2017

$f ' \left(x\right) = - \frac{2 {x}^{2} - 4}{4 - {x}^{2}} ^ \left(\frac{1}{2}\right)$

Explanation:

$\text{given "f(x)=g(x)h(x)" then }$

$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right) \leftarrow \text{ product rule}$

$g \left(x\right) = x \Rightarrow g ' \left(x\right) = 1$

$h \left(x\right) = {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} \leftarrow \text{ differentiate using "color(blue)"chain rule}$

$\Rightarrow h ' \left(x\right) = \frac{1}{2} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \times \frac{d}{\mathrm{dx}} \left(4 - {x}^{2}\right)$

$\textcolor{w h i t e}{\Rightarrow h ' \left(x\right)} = - x {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}}$

$\Rightarrow f ' \left(x\right) = x \left(- x {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}}\right) + {\left(4 - {x}^{2}\right)}^{\frac{1}{2}}$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- {x}^{2} + 4 - {x}^{2}\right)$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = - \frac{2 {x}^{2} - 4}{4 - {x}^{2}} ^ \left(\frac{1}{2}\right)$