Question

How do you find the derivative (quotient rule) for `(x^2 + 8x + 3)/sqrtx`?

Answer 1

`y^' = 1/2 * x^(-3/2) * (3x^2 + 8x - 3)`

Explanation:

The quotient rule tells you that you can differentiate functions that can be written as the quotient of two other functions, let's say `f(x)` and `g(x)` by using the formula

`color(blue)(d/dx(y) = ([d/dx(f(x))] * g(x) - f(x) * d/dx(g(x)))/(g(x))^2`, with `g(x)!=0`

In your case, you have `f(x) = x^2 + 8x + 3` and `g(x) = sqrt(x)`. The derivative of `y` will thus be

`d/dx(y) = ([d/dx(x^2 + 8x + 3)] * sqrt(x) - (x^2 + 8x + 3) * d/dx(sqrt(x)))/(sqrt(x))^2`

`y^' = ((2x + 8) * sqrt(x) - (x^2 + 8x + 3) * 1/2 * 1/sqrt(x))/x`

`y^' = ((2x + 8) * x^(1/2) - 1/2 * (x^2 + 8x + 3) * x^(-1/2))/x`

This is equivalent to

`y^' = (x^(-1/2) * [(2x + 8) * x - 1/2(x^2 + 8x + 3)])/x`

`y^' = (x^(-1/2) * (2x^2 + 8x -1/2x^2 -4x -3/2))/x`

`y^' = 1/2 * x^(-1/2)/x * (4x^2 + 16x -x^2 - 8x - 3)`

`y^' = color(green)(1/2 * x^(-3/2) * (3x^2 + 8x - 3))`