Calculate the Derivative

[irenevecchio]

Could I get some help with this please. I have this as a homework problem and I am not getting any of the answers I have to choose from. I came close, but not quite what is written. My answer to this is 2/cube rt of 9?
Given f(u) =cube root of u and g(x) = u = 1 + 2x3, find (f◦g)'(0).

 

[JeffM]

Could I get some help with this please. I have this as a homework problem and I am not getting any of the answers I have to choose from. I came close, but not quite what is written. My answer to this is 2/cube rt of 9?
Given f(u) =cube root of u and g(x) = u = 1 + 2x3, find (f◦g)'(0).

The derivative of a composition of functions equals the product of the derivatives of the constituent functions. This is called the CHAIN RULE, and it is very important. So let's take this step by step because you must get very comfortable with this rule to do well in calculus.

$\displaystyle f(u) = \sqrt[3]{u} = u^{(1/3)}\ and\ g(x) = u = 1 + 2x^3.$ Are those what is given? If so, you should have written g(x) = 1 + 2x^3.

So what does f(g(x)) even mean? It means $\displaystyle f(g(x)) = f(1 + 2x^3) = \sqrt[3]{1 + 2x^3} = h(x) = y.$

A composition of functions is just another function. I have shown that above by saying f(g(x)) = h(x) = y.

Now I like to use the old-fashioned notation for chain rule problems because it contains a sort of mnemonic for the process.

The derivative of y = h(x) is $\displaystyle \dfrac{dy}{dx}.$ The chain rule says: $\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx}.$

Stated that way, the rule looks similar to what you learned in grade school about cancelling when multiplying fractions. (Being rigorous, this is just a seeming because a derivative is a limit, not a fraction.) I like it, however, because it reminds you of something you learned long ago. Are you with me so far?

$\displaystyle y = u^{(1/3)} \implies \dfrac{dy}{du} = what?$ The answer will be in terms of u.

$\displaystyle u = 1 + 2x^3 \implies \dfrac{du}{dx} = what?$ The answer will be in terms of x.

But it is possible to substitute an expression in terms of x for u so, in terms of x, $\displaystyle \dfrac{dy}{du} = what?$

Now you have both derivatives in terms of x so $\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = what$ in terms of x?

So if x = 0, what is $\displaystyle \dfrac{dy}{dx}?$