Question

How do you use the chain Rule to find the derivative of `sqrt(2x^3 - 3x- 4)`?

Answer 1

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

Explanation:

You can use the chain rule to find the derivative of this function by using `u = 2x^3 - 3x - 4`.

The chain rule tells you that you can differentiate a function `y` that depends on a variable `u`, which in turn depends on another variable `x`, by

`color(blue)(d/dx(y) = d/(du)(y) * d/dx(u))`

Two additional rules of derivation, the power rule and the sum rule will come into play at certain points in the derivation of `y`.

So, if you use `u = 2x^3 - 3x - 4`, then you have

`y = sqrt(u) = u^(1/2)`

So, the derivative of `y` will look like this

`y^' = d/(du)(y) * d/dx(u)`

`y^' = d/(du)(u^(1/2)) * d/dx(2x^3 - 3x - 4)`

The power rule tells you that you can differentiate a variable `x` raised to a power `a` by

`color(blue)(x^a = a * x^(a-1))`

This means that you have

`d/(du)(u^(1/2)) = 1/2 * u^(1/2 - 1)`

`d/(du)(u^(1/2)) = 1/2 * 1/u^(1/2) = 1/2 * 1/sqrt(u)`

The sum rule tells you that the derivative of a function that can be written as a sum of two (or more) functions is equal to the sum of the derivatives of those functions.

For `color(blue)(y = f(x) + g(x) + h(x) + ...)`

you have

`color(blue)(d/dx(y) = f^'(x) + g^'(x) + h^'(x) + ...)`

This means that you can write

`d/dx(2x^3 - 3x - 4) = d/dx(2x^3) + d/dx(-3x) + d/dx(-4)`

`d/dx(2x^3 - 3x - 4) = 2 * 3 x^(3-2) -3 + 0`

`d/dx(2x^3 - 3x - 4) = 6x^2 - 3`

Your original derivative now becomes

`y^' = 1/2 * 1/sqrt(u) * (6x^2 - 3)`

`y^] = 1/2 * 1/sqrt((2x^3 - 3x - 4)) * 3(2x^2 - 1)`

Finally,

`y^' = color(green)(3/2 * (2x^2 - 1)/sqrt(2x^3 - 3x - 4))`