Video Transcript
Evaluate the first derivative of the function 𝑦 equals four 𝑥 multiplied by the square root of 𝑥 plus six at the point with coordinates negative two, negative 16.
Let’s have a look at this function 𝑦. We can see that it consists of a product of terms. It’s equal to four 𝑥 multiplied by the square root of 𝑥 plus six. And as the exponent in that second factor is a half, we can’t simply distribute the parentheses in order to turn this into a polynomial function.
Instead, we’re going to need to use the product rule of differentiation. This tells us that, for two differentiable functions 𝑢 and 𝑣, the derivative with respect to 𝑥 of their product 𝑢𝑣 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. We multiply each function by the derivative of the other and add these together. So we define 𝑢 to be one of our factors. We’ll let 𝑢 equal four 𝑥. And we define 𝑣 to be the other, the square root of 𝑥 plus six, which we’ve already said can be expressed as 𝑥 plus six to the power of one-half.
We then need to find each of their individual derivatives. d𝑢 by d𝑥 is straight forward. We can use the power rule of differentiation. We can think of four 𝑥 as four 𝑥 to the first power. So in order to differentiate, we multiply by the exponent of one and then reduce the exponent by one, giving four multiplied by one multiplied by 𝑥 to the power of zero. 𝑥 to the power of zero is simply one. So we have four multiplied by one multiplied by one, which is just four.
Now, d𝑣 by d𝑥 is a little bit more complicated because we see that 𝑣 is actually a composite function. It’s a function of a function. We take the function 𝑥 plus six and then find its square root. In order to find this derivative then, we’re going to need to use the chain rule. The chain rule tells us that if 𝑣 is a function of some other function 𝑧, which is itself a function of 𝑥, then the derivative of 𝑣 with respect to 𝑥 is equal to the derivative of 𝑣 with respect to 𝑧 multiplied by the derivative of 𝑧 with respect to 𝑥. I’ve changed the letters from what you might usually see here. Often, we see it written in terms of 𝑢, because we already have 𝑢 defined to be another function in this question.
So, we can introduce this new variable 𝑧. We’ll let 𝑧 equal 𝑥 plus six. That’s the inner part of our composite function. 𝑣 will then be equal to 𝑧 to the power of one-half. So 𝑣 is a function of 𝑧. And 𝑧 is a function of 𝑥. We then find each of their individual derivatives, both of which we can do using the power rule of differentiation. The derivative of 𝑧 with respect to 𝑥 is simply one. And the derivative of 𝑣 with respect to 𝑧 is a half multiplied by 𝑧 to the power of negative one-half.
Applying the chain rule then, we have that d𝑣 by d𝑥 is equal to d𝑣 by d𝑧 multiplied by d𝑧 by d𝑥. That’s one-half 𝑧 to the power of negative one-half multiplied by one. But of course, multiplying by one has no effect. We can rewrite this as one over two 𝑧 to the power of a half. And finally, we must remember to undo our substitution. So we need to give d𝑣 by d𝑥 in terms of 𝑥. We find then that d𝑣 by d𝑥 is equal to one over two multiplied by 𝑥 plus six to the power of a half. We’ve replaced 𝑧 with 𝑥 plus six.
Now, you may also be able to find this derivative using the general power rule of differentiation rather than the chain rule. Which tells us that if we have some function 𝑓 of 𝑥 to the 𝑛th power, then its derivative with respect to 𝑥 is equal to 𝑛 multiplied by 𝑓 prime of 𝑥 multiplied by 𝑓 of 𝑥 to the power of 𝑛 minus one. It’s a lot like the power rule of differentiation. But we have this extra factor 𝑓 prime of 𝑥. That’s the derivative of our function 𝑓 of 𝑥.
For our function 𝑣 then, the function 𝑓 of 𝑥 would be 𝑥 plus six. So applying the general power rule, we’d find that d𝑣 by d𝑥 is equal to 𝑛, that’s one-half, multiplied by 𝑓 prime of 𝑥, that’s the derivative of 𝑥 plus six, which is simply one, multiplied by 𝑓 of 𝑥 to the power of 𝑛 minus one. That’s 𝑥 plus six to the power of negative one-half. This can of course be written as one over two multiplied by 𝑥 plus six to the power of a half, which is what we’ve already found for our derivative of 𝑣 with respect to 𝑥 using the chain rule. So either of these two methods would be absolutely fine.
In either case, now that we have both d𝑢 by d𝑥 and d𝑣 by d𝑥, we need to return to our formula for the product rule. We have that d𝑦 by d𝑥 is equal to 𝑢 times d𝑣 by d𝑥. That’s four 𝑥 multiplied by one over two 𝑥 plus six to the power of a half, which we can write as four 𝑥 over two 𝑥 plus six to the power of a half. Plus 𝑣 times d𝑢 by d𝑥. That’s four multiplied by 𝑥 plus six to the power of one-half. We can simplify by a factor of two in our first term. And we may also find it helpful to rewrite those exponents of one-half as square roots. So we have that d𝑦 by d𝑥 is equal to two 𝑥 over the square root of 𝑥 plus six plus four multiplied by the square root of 𝑥 plus six.
Finally, we recall that we weren’t asked to give a general expression for the first derivative. We were asked to evaluate it at a particular point. So the final step is to substitute the 𝑥-coordinate of this point, that’s negative two, into our expression for d𝑦 by d𝑥. Doing so gives d𝑦 by d𝑥 is equal to two multiplied by negative two over the square root of negative two plus six plus four multiplied by the square root of negative two plus six. That simplifies to negative four over the square root of four plus four multiplied by the square root of four. Now, the square root of four is of course two. So we have negative four over two plus four multiplied by two. That’s negative two plus eight, which is equal to six.
So in this question, we’ve had to use more than one rule of differentiation, either the product rule and the formal chain rule or the product rule and the general power rule. But in doing so, we’ve found that the first derivative of the function 𝑦 equals four 𝑥 multiplied by the square root of 𝑥 plus six at the point with coordinates negative two, negative 16 is six.
