How do you differentiate f(x)= (4x^5+5)^(1/2) using the chain rule?

May 22, 2018

$\frac{d \left[f \left(x\right)\right]}{\mathrm{dx}} = \frac{10 {x}^{4}}{\sqrt{4 {x}^{5} + 5}}$

Explanation:

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

$\frac{d \left[f \left(x\right)\right]}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[{\left(4 {x}^{5} + 5\right)}^{\frac{1}{2}}\right]$

let $u = 4 {x}^{5} + 5$

$\frac{d \left[f \left(x\right)\right]}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \left[{u}^{\frac{1}{2}}\right] \frac{d}{\mathrm{dx}} \left[4 {x}^{5} + 5\right]$

$\frac{d \left[f \left(x\right)\right]}{\mathrm{dx}} = \frac{1}{2} {u}^{- \frac{1}{2}} \times 20 {x}^{4}$

Substitute $u$ back in

$\frac{d \left[f \left(x\right)\right]}{\mathrm{dx}} = \frac{1}{2} {\left(4 {x}^{5} + 5\right)}^{- \frac{1}{2}} \times 20 {x}^{4}$

Simplify

$\frac{d \left[f \left(x\right)\right]}{\mathrm{dx}} = \frac{10 {x}^{4}}{\sqrt{4 {x}^{5} + 5}}$

May 22, 2018

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

Explanation:

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

$f ' \left(x\right) = g ' \left(h \left(x\right)\right) \times h ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$\Rightarrow f ' \left(x\right) = \frac{1}{2} {\left(4 {x}^{5} + 5\right)}^{- \frac{1}{2}} \times \frac{d}{\mathrm{dx}} \left(4 {x}^{5} + 5\right)$

"color(white)(rArrf'(x))=1/cancel(2)(4x^5+5)^(-1/2)xxcancel(20)^(10)x^4

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \frac{10 {x}^{4}}{4 {x}^{5} + 5} ^ \left(\frac{1}{2}\right)$