### Video Transcript

Find ππ¦ by ππ₯ at π₯ equals negative two when π¦ is equal to four π₯ minus one times negative three π₯ cubed plus seven.

Weβre looking for ππ¦ by ππ₯, which is the derivative of π¦ with respect to π₯. And so we need to differentiate four π₯ minus one times negative three π₯ cubed plus seven with respect to π₯. Iβd like to show two methods for solving this problem in this video. In the first method, weβll expand the right-hand side using the distributive property and then differentiate term by term. And the second method weβll see uses the product rule for differentiation, which allows us to find the derivative of a product of factors as long as we know the derivatives of those factors.

More on that later, first letβs expand using the distributive property. π¦ is now written as a polynomial in π₯, which we can differentiate term by term. We differentiate both sides with respect to π₯. And as claimed, we can differentiate each term individually and then add or subtract the results as appropriate.

Now, how do we differentiate each term? Well, we have a formula for the derivative of a power of π₯ with respect to π₯: π by ππ₯ of π₯ to the π is π times π₯ to the π minus one. In other words, to differentiate a power of π₯ with respect to π₯, you take a copy of the exponents down to the front multiplying by it and then you reduce the exponent thatβs left by one.

And as the derivative of some number times a function is just that number times the derivative of the function, the derivative π by ππ₯ of π times π₯ to the π is π times π π₯ to the π minus one. We can apply this rule to find π by ππ₯ of negative 12π₯ to the four with π equal to negative 12 and π₯ equal to four. We get negative 12 times four π₯ to the four minus one which is negative 48 π₯ cubed.

Moving on, the derivative of 28π₯ with respect to π₯ is just 28 because the derivative of π₯ with respect to π₯ is just one. That something you can work out using the power rule setting π equal to one or just work out by knowing something about straight line graphs.

The derivative of three π₯ to the three with respect to π₯ is nine π₯ squared, where the nine comes from multiplying the coefficient three by the exponent three. And having multiplied, we need to reduce the exponent by one β from three to two.

Finally, the derivative of the constant function seven is just zero. So this is what we get for ππ¦ by ππ₯. If this is what weβre looking for, then we probably want to swap the two last terms nine π₯ squared and 28 so the terms are written in order of degree. But actually, weβre not interested in the derivative in terms of π₯. Apart from that, it allows us to find the derivative when π₯ is equal to negative two.

So we have to evaluate this expression at π₯ equals negative two. Evaluating this using a calculator, we get 448. We found this value by expanding the expression for π¦ in terms of π₯ using the distributive property and then differentiating the polynomial that we got. But we can also solve this problem using the product rule.

The product rule states that the derivative of the product of two functions is equal to the first function times the derivative of the second function plus the derivative of the first function times the second function. We can apply the product rule to find the derivative of the product of four π₯ minus one and negative three π₯ cubed plus seven.

With π of π₯ equal to four π₯ minus one and π of π₯ equal to negative three π₯ cubed plus seven, the first term on the right-hand side π of π₯ times the derivative of π of π₯ becomes four π₯ minus one times the derivative of negative three π₯ cubed plus seven. And the second term becomes the derivative of four π₯ minus one times negative three π₯ cubed plus seven.

Our task now is to find these two derivatives. And we do this using the rules from earlier in the video. In particular, we need the derivative of a number times a power of π₯ with respect to π₯. Using this rule and the fact that the derivative of the constant function seven is zero, we find that the first derivative is negative nine π₯ squared. We find out that derivative of four π₯ minus one with respect to π₯ is four either by using our rule or using what we know about straight lines. We can substitute these in, giving us ππ¦ by ππ₯ in terms of π₯. And then, all thatβs left to do is to substitute negative two for π₯. And doing so, we get 448 as before.

Here weβve seen two methods for finding the value of ππ¦ by ππ₯ at π₯ equals negative two. Of course, there are other methods we couldβve used. We couldβve derived this value from our first principles using the definition of a derivative in terms of limits.

The fact that the product rule gives the same answer as expanding and differentiating term by term is encouraging because it suggests that the product rule that we have is actually right. In a similar way, you can check that someone who thinks that the derivative of a product of functions is just the product of their derivatives will get the answer wrong. If you ever concerned that youβve remembered the product rule incorrectly, then you can check that it works on a simple example like this.

In this problem, weβve seen that we donβt need to use the product rule to find the answer. But in other problems, there will be no way around using the product rule. Weβll have to use it and so weβll have to remember it.