Question

How do you find the derivative of `f(x)=sqrt(ax+b)`?

Answer 1

`f'(x)=a/(2sqrt(ax+b))`

Explanation:

differentiate using the `color(blue)"chain rule"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(2/2)|)))to(A)`

`"let " u=ax+brArr(du)/(dx)=a`

`rArry=u^(1/2)rArr(dy)/(du)=1/2u^(-1/2)`

substitute into (A), changing u back into terms of x

`rArrdy/dx=1/2u^(-1/2).a=a/(2sqrt(ax+b))`