Question

How do you differentiate `f(x)=(x^3-2x+3)^(3/2)` using the chain rule?

Answer 1

`3/2 * (sqrt(x^3 - 2x + 3)) * (3x^2 - 2)`

Explanation:

The chain rule:
`d/dx f(g(x)) = f'(g(x)) * g'(x)`

The power rule:
`d/dx x^n = n*x^(n-1)`

Applying these rules:
1 The inner function, `g(x)` is `x^3-2x+3`, the outer function, `f(x)` is `g(x)^(3/2)`

2 Take the derivative of the outer function using the power rule
`d/dx (g(x))^(3/2) = 3/2 * g(x)^(3/2 - 2/2) = 3/2 * g(x)^(1/2) = 3/2 * sqrt(g(x))`
`f'(g(x)) = 3/2 * sqrt(x^3 - 2x + 3)`

3 Take the derivative of the inner function
`d/dx g(x) = 3x^2 -2`
`g'(x) = 3x^2 -2`

4 Multiply `f'(g(x))` with `g'(x)`
`(3/2 * sqrt(x^3 - 2x + 3) ) * (3x^2 - 2)`

solution: `3/2 * (sqrt(x^3 - 2x + 3)) * (3x^2 - 2)`