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Write the composite function in the form $ f(g(x)…

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Problem 1 Medium Difficulty

Write the composite function in the form $ f(g(x)). $ [Identify the inner function $ u = g(x) $ and the outer function $ y = f(u). $ ] Then find the derivative $ dy/ dx. $
$ y = \sqrt[3]{1 + 4x} $

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Video Transcript

all right. So for this function, the Cuba symbol is serving as a grouping symbol. And so, inside the cube root, we find our inside function. So we'll call that G of X g of X equals one plus four X. That's our inside function. And then the outside function is the cube root function. And we're going to call that f of X so f of X is the cube root of X. So now we're ready to use the chain rule to find the derivative of this function. And the derivative is D Y D X, and that's going to be the derivative of the inside multiplied by the derivative of the outside. Now for the cube root function, I like to rename it as X to the 1/3 power. And so I'm going to rename my function before I start. Rewrite it, I should say, as why equals one plus four x to the 1/3 power. Okay, so now for the derivative D Y d x, the derivative of the outside. So we would bring down the 1/3 power and then raise its inside 21 less power, so that would be the negative 2/3 power. Now we're going to multiply by the derivative of the inside. So the derivative of one plus four X is for now. We just want to simplify the answer. So let's multiply the four and the 1/3 when we get 4/3 times one plus four x to the negative 2/3 power. Now, you might just leave your answer like that. However, if you've been instructed to simplify, what you might want to do is eliminate your negative exponents by taking that term to the denominator so you would have four over three times the quantity one plus four x two the 2/3 power. Or you might want to eliminate the exponents form and change it back to radical forms, since the original problem was in radical form. So your answer would look like for over three times the cube root of one plus four x squared

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Calculus: Early Transcendentals

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