### Video Transcript

If π of π₯ equals three π₯ to the fourth power plus 16 times the cube root of π₯ plus four π₯, find π prime of π₯.

π of π₯ is itself the product of two differentiable functions, the first one being three π₯ to the fourth power plus 16 and the second being the cube root of π₯ plus four π₯. This means we can use the product rule to find its derivative, π prime of π₯.

This says that the derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. Weβll define π’ as our first function, three π₯ to the fourth power plus 16. Then, weβll define π£ as our second function, changing the cube root of π₯ to π₯ the power of one-third. So, π£ is equal to π₯ to the power of one-third plus four π₯.

We can see from the formula for the product rule that weβre going to need to differentiate each of these with respect to π₯. And we recall that to differentiate a polynomial term, as all of these are, we multiply the entire term by the exponent and then simply reduce that exponent by one. The derivative, then, of three π₯ to the fourth power is four times three π₯ cubed. And the derivative of 16, a constant, is just zero.

So, dπ’ by dπ₯ simplifies to 12π₯ cubed. The derivative of π₯ to the power of one-third is a third times π₯ to the power of negative two-thirds. And then, the derivative of four π₯ is simply four. Letβs replace π’, dπ£ by dπ₯, π£, and dπ’ by dπ₯ in our formula for the derivative of π’ times π£. Itβs π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.

We then convert each expression with a fractional index back into root four. And we see that π₯ to the power of negative two-thirds is one over the cube root of π₯ squared, whereas π₯ to the power of one-third is simply the cube root of π₯. And in doing so, we found π prime of π₯. Itβs one over three times the cube root of π₯ squared plus four times three π₯ to the fourth power plus 16 plus 12π₯ cubed times four π₯ plus the cube root of π₯.