Question

How do you find the derivative for `sqrt(x)/(x^3+1)`?

Answer 1

`y^' = (1-5x^3)/(2sqrt(x) * (x^3+1)^2)`

Explanation:

You can differentiate this function by using the quotient rule, which allows you to differentiate a function that can be written as

`color(blue)(y = f(x)/g(x))`, with `g(x) !=0`

by using the formula

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

In your case, you can say that

`f(x) = sqrt(x)" "` and `" "g(x) = x^3 + 1`

This means that you can write

`d/dx(y) = ([d/dx(sqrt(x))] * (x^3 + 1) - sqrt(x) * d/dx(x^3 - 1))/(x^3 + 1)^2`

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

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

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

`y^' = 1/2 * x^(-1/2) * (1-5x^3)/(x^3 + 1)^2`

This can be rewritten as

`y^' = color(green)((1-5x^3)/(2sqrt(x) * (x^3+1)^2))`