Question

How do you differentiate `y=sqrt(4+3x)`?

Answer 1

`y=3/(2(4+3x)^(1/2))`

Explanation:

Recall the formula for chain rule:

`color(blue)(bar(ul(|color(white)(a/a)dy/dx=dy/(du)(du)/dxcolor(white)(a/a)|)))` or `color(blue)(bar(ul(|color(white)(a/a)f'(x)=g'[h(x)]h'(x)color(white)(a/a)|)))`

and the formula for power rule:

`color(blue)(bar(ul(|color(white)(a/a)d/dx(x^n)=nx^(n-1)color(white)(a/a)|)))`

To start, recognize the inside and outside functions of `y=color(green)(sqrt(color(darkorange)(4+3x)))`.

Inside function: `y=color(darkorange)(4+3x)`
Outside function: `y=color(green)(sqrt(a))`

How to Differentiate Using Chain Rule

`1`. Take the derivative of the outside function, `y=sqrt(a)`, but replace the `a` with the inside function, `4+3x`.

`2`. Multiply by the derivative of the inside function, `4+3x`.

Applying Chain Rule
1. The derivative of the outside function, `y=sqrt(a)`, would be `1/2a^(-1/2)`, using the power rule. However, `a` needs to be replaced by the inside function, so it becomes `1/2(4+3x)^(-1/2)`.

`y=sqrt(a)`

`color(red)(darr)`

`y=1/2a^(-1/2)`

`color(red)(darr)`

`y=1/2(4+3x)^(-1/2)`

2.`color(white)(i)`Then we need to multiply by the derivative of the inside function, `4+3x`, which becomes `3` using the power rule.

`y=1/2(4+3x)^(-1/2)`

`color(red)(darr)`

`y=1/2(4+3x)^(-1/2)(3)`

`color(green)(bar(ul(|color(white)(a/a)y=3/(2(4+3x)^(1/2))color(white)(a/a)|)))`