Question

How to find derivative of this
`x√(4-x)`

Answer 1

We can use the product rule

G(x)=f(x)*h(x)

G'(x)=f'(x)h(x) +f(x)h'(x)

`=1√(4-x) + x(1)/(2√(4-x) ) *-1\\=√(4-x) +(-x)/(2√(4-x) )`

I hope this helps

Answer 2

`\frac{\text{d}y}{\text{d}x} = (-3x+8)/(2√(4-x))`

Explanation:

`\boxed{\begin{minipage}{5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\frac{\text{d}y}{\text{d}x}=u\frac{\text{d}v}{\text{d}x}+v\frac{\text{d}u}{\text{d}x}$\\\end{minipage}}`

Given equation:

`y=x√(4-x)`

Identify u and v:

`\textsf{Let $u=x$}`

`\textsf{Let $v=√(4-x)=(4-x)^{(1)/(2)}$}`

Differentiate u and v with respect to x.

Differentiate u:

`u=x \implies \frac{\text{d}u}{\text{d}x}=1`

Differentiate v using the chain rule:

`\begin{aligned}v=(4-x)^{(1)/(2)} \implies \frac{\text{d}v}{\text{d}x} &=(1)/(2)(4-x)^{((1)/(2)-1)} \cdot -1 \\\\ & =-(1)/(2)(4-x)^{-(1)/(2)}\\\\&=-(1)/(2√(4-x))\end{aligned}`

Put everything into the product rule formula:

`\begin{aligned}\implies \frac{\text{d}y}{\text{d}x} & =u\frac{\text{d}v}{\text{d}x}+v\frac{\text{d}u}{\text{d}x}\\\\& = x \cdot \left(-(1)/(2√(4-x))\right)+√(4-x) \cdot 1\\\\& = -(x)/(2√(4-x))+√(4-x)\\\\& = -(x)/(2√(4-x))+(2√(4-x)√(4-x))/(2√(4-x))\\\\& = -(x)/(2√(4-x))+(2(4-x))/(2√(4-x))\\\\& = (-x+8-2x)/(2√(4-x))\\\\& = (-3x+8)/(2√(4-x))\end{aligned}`

Differentiation Rules

`\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}u}*\frac{\text{d}u}{\text{d}x}$\\\end{minipage}}`

`\boxed{\begin{minipage}{4.5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\frac{\text{d}y}{\text{d}x}=nx^(n-1)$\\\end{minipage}}`

`\boxed{\begin{minipage}{4.5 cm}\underline{Differentiating $ax$}\\\\If $y=ax$, then $\frac{\text{d}y}{\text{d}x}=a$\\\end{minipage}}`