Video: Differentiating a Combination of Polynomial Functions Using the Product Rule

Evaluate d𝑦/dπ‘₯ at π‘₯ = 1 if 𝑦 = (6π‘₯Β² βˆ’ 2π‘₯ βˆ’ 3)⁻³ (π‘₯Β² βˆ’ 2)⁡.

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Video Transcript

Evaluate d𝑦 by dπ‘₯ at π‘₯ equals one if 𝑦 equals six π‘₯ squared minus two π‘₯ minus three to the power of negative three multiplied by π‘₯ squared minus two to the power of five.

So the first thing we need to do is find d𝑦 by dπ‘₯ by differentiating our function. And whatever we get, we’ll then evaluate it at π‘₯ equals one. So we want to differentiate our function 𝑦. And we can see that 𝑦 is actually two functions being multiplied together. It’s a product, so we use the product rule. Let’s recall the product rule. The product rule says that if 𝑦 equals 𝑒𝑣, then d𝑦 by dπ‘₯ equals 𝑒 d𝑣 dπ‘₯ plus 𝑣 d𝑒 dπ‘₯.

So, for our question, our 𝑒 will be six π‘₯ squared minus two π‘₯ minus three to the power of negative three. And our 𝑣 will be π‘₯ squared minus two to the power of five. And we can see from the product rule we’re going to need d𝑣 by dπ‘₯ and d𝑒 by dπ‘₯. So let’s go ahead and find d𝑒 by dπ‘₯. Now because 𝑒 is a function of a function, we’re going to need the chain rule. So let’s just remember the chain rule first of all. The chain rule says that if 𝑒 is 𝑓 of π‘Ž and π‘Ž is 𝑔 of π‘₯, then d𝑒 by dπ‘₯ is equal to d𝑒 by dπ‘Ž multiplied by dπ‘Ž by dπ‘₯. So let’s apply this to our function 𝑒.

So we have that π‘Ž is six π‘₯ squared minus two π‘₯ minus three. And that differentiates with respect to π‘₯ to get that dπ‘Ž by dπ‘₯ is equal to 12π‘₯ minus two. And we got that by multiplying the coefficient six by the power two and then taking one away from the power to get 12π‘₯. The two π‘₯ differentiates to give us two. And then the three is just a constant. And because we said that π‘Ž is equal to six π‘₯ squared minus two π‘₯ minus three, we have that 𝑒 is equal to π‘Ž to the power of negative three. And so d𝑒 by dπ‘Ž is negative three π‘Ž to the power of negative four.

And so, applying the formula for the chain rule, we have that d𝑒 by dπ‘₯ is equal to negative three π‘Ž to the power of negative four multiplied by 12π‘₯ minus two. And remember, π‘Ž, we already defined to be six π‘₯ squared minus two π‘₯ minus three. So by replacing π‘Ž, we get negative three multiplied by six π‘₯ squared minus two π‘₯ minus three to the power of negative four multiplied by 12π‘₯ minus two. So that gives us d𝑒 by dπ‘₯. And now we also need d𝑣 by dπ‘₯. And, again, 𝑣 is a function of a function. So we’re going to need the chain rule again.

So, here, we have that π‘Ž is π‘₯ squared minus two, so that dπ‘Ž by dπ‘₯ is two π‘₯. As we defined π‘Ž to be π‘₯ squared minus two, we have that 𝑣 is equal to π‘Ž to the power of five so that d𝑣 by dπ‘Ž is equal to five π‘Ž to the power of four. And then applying the chain rule gives us that d𝑣 by dπ‘₯ is equal to five π‘Ž to the power of four multiplied by two π‘₯. And remember, again, we’re to find π‘Ž to be π‘₯ squared minus two. So we can replace π‘Ž in this expression, which gives us five multiplied by π‘₯ squared minus two to the power of four multiplied by two π‘₯. And we can multiply together the five and the two π‘₯ to tidy up this a little bit. So that gives us 10π‘₯ multiplied by π‘₯ squared minus two to the power of four.

So now we have d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯. So we can apply the product rule. So I’m just going to clear some space here. Okay, so applying the product rule gives us this result. So this is d𝑦 by dπ‘₯. But, remember, in the question, we were asked to evaluate d𝑦 by dπ‘₯ at π‘₯ equals one. This means we need to substitute in π‘₯ equals one into our first derivative. And once we’ve substituted in one, we can tidy up our answer a little bit. And one to the power of negative three is just one. And negative one to the power of four is also one.

So this is just one times 10 times one. So that gives us 10. And negative one to the power of five is negative one. And one to the power of negative four is just one. So this is negative one times negative three times 10, which is 30. So we’ve got 10 plus 30, which gives us 40. So we differentiated our function using a combination of the chain rule and the product rule. And then we evaluated our first derivative at the point π‘₯ equals one to get our final answer of 40.

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