Question Video: Differentiating Composite Functions Involving Rational and Root Functions Using the Chain Rule | Nagwa Question Video: Differentiating Composite Functions Involving Rational and Root Functions Using the Chain Rule | Nagwa

Question Video: Differentiating Composite Functions Involving Rational and Root Functions Using the Chain Rule Mathematics • Second Year of Secondary School

Given that 𝑦 = (𝑧²)/(3 + 9𝑧²) and 𝑧 = √(8𝑥 + 3), determine d𝑦/d𝑥.

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Video Transcript

Given that 𝑦 is equal to 𝑧 squared divided by three plus nine 𝑧 squared and 𝑧 is equal to the square root of eight 𝑥 plus three, determine d𝑦 by d𝑥.

We need to find d𝑦 by d𝑥. That’s the first derivative of 𝑦 with respect to 𝑥. But we’re not given 𝑦 as a function in 𝑥. Instead, we’re given 𝑦 as a function in 𝑧, and we can see that 𝑧 is also given as a function in 𝑥. This means we’ll need to find d𝑦 by d𝑥 by using the chain rule. Recall, this tells us if 𝑦 is a function in 𝑧 and 𝑧 in turn is a function in 𝑥, then d𝑦 by d𝑥 will be equal to d𝑦 by d𝑧 times d𝑧 by d𝑥. So now we need to differentiate 𝑦 with respect to 𝑧 and then 𝑧 with respect to 𝑥.

Let’s start with differentiating 𝑦 with respect to 𝑧. We can see this is the quotient of two polynomials. So we’ll do this by using the quotient rule. So we’ll start by recalling the quotient rule. If 𝑢 and 𝑣 are differentiable functions, then the derivative of 𝑢 of 𝑧 divided by 𝑣 of 𝑧 with respect to 𝑧 is equal to 𝑢 prime of 𝑧 times 𝑣 of 𝑧 minus 𝑣 prime of 𝑧 times 𝑢 of 𝑧 all divided by 𝑣 of 𝑧 all squared. So to find d𝑦 by d𝑧, we’ll set 𝑢 of 𝑧 to be the function in our numerator, that’s 𝑧 squared, and 𝑣 of 𝑧 to be the function in our denominator, that’s three plus nine 𝑧 squared.

And to apply the quotient rule, we’re going to need to find expressions for 𝑢 prime of 𝑧 and 𝑣 prime of 𝑧. We can do both of these by using the power rule for differentiation. We see that 𝑢 prime of 𝑧 will be equal to two 𝑧 and 𝑣 prime of 𝑧 will be equal to 18𝑧. Now that we found expressions for 𝑢 prime and 𝑣 prime, we can find d𝑦 by d𝑧 by using the quotient rule. Substituting in our expressions for 𝑢, 𝑣, 𝑢 prime, and 𝑣 prime, we get two 𝑧 times three plus nine 𝑧 squared minus 18𝑧 times 𝑧 squared all divided by three plus nine 𝑧 squared all squared.

And we can simplify this expression. In our numerator, if we distribute over the parentheses, one of our terms will be 18𝑧 cubed. However, we also have in our numerator negative 18𝑧 cubed. So all that’s left in our numerator is six 𝑧. Therefore, d𝑦 by d𝑧 is equal to six 𝑧 divided by three plus nine 𝑧 squared all squared.

So we found an expression for d𝑦 by d𝑧 . All that’s left now is to find d𝑧 by d𝑥. So we want to find d𝑧 by d𝑥, and 𝑧 is equal to the square root of eight 𝑥 plus three. There’s a few different ways we could differentiate this. We could use the chain rule or we could use the general power rule. We can use either method; we’re going to use the general power rule. First, we’ll rewrite 𝑧 as eight 𝑥 plus three all raised to the power of one-half. Next, we need to recall the general power rule. This tells us for differentiable function 𝑓 of 𝑥 and any constant 𝑛, the derivative of 𝑓 of 𝑥 all raised to the 𝑛th power with respect to 𝑥 is equal to 𝑛 times 𝑓 prime of 𝑥 multiplied by 𝑓 of 𝑥 raised to the power of 𝑛 minus one.

In our case, our exponent 𝑛 is one-half and our inner function 𝑓 of 𝑥 is eight 𝑥 plus three. And if 𝑓 of 𝑥 is the linear function eight 𝑥 plus three, then 𝑓 prime of 𝑥 will be the slope of this function, which is eight. We’re now ready to find an expression for d𝑧 by d𝑥 by using the general power rule. Our value of 𝑛 is one-half and 𝑓 prime of 𝑥 is equal to eight. So we get d𝑧 by d𝑥 is equal to one-half times eight multiplied by eight 𝑥 plus three all raised to the power of negative one-half. And we can simplify this. First, one-half multiplied by eight is equal to four. Next, eight 𝑥 plus three all raised to the power of negative one-half is the same as dividing by the square root of eight 𝑥 plus three.

So we found an expression for d𝑧 by d𝑥. It’s equal to four divided by the square root of eight 𝑥 plus three. So we found an expression for d𝑦 by d𝑧 and d𝑧 by d𝑥. This means we can use the chain rule to find an expression for d𝑦 by d𝑥. So by substituting in our expressions for d𝑦 by d𝑧 and d𝑧 by d𝑥, we get d𝑦 by d𝑥 is equal to four divided by the square root of eight 𝑥 plus three multiplied by six 𝑧 divided by three plus nine 𝑧 squared all squared. And at this point, we can simplify. Remember, in the question, we’re told that 𝑧 is equal to the square root of eight 𝑥 plus three, so we could start substituting this into our expression.

However, notice in our numerator, we have 𝑧 and in our denominator, we have the square root of eight 𝑥 plus three, which is equal to 𝑧. So we can cancel this shared factor of 𝑧 in our numerator and denominator. This gives us 24 divided by three plus nine 𝑧 squared all squared. Now we’ll use our substitution 𝑧 is equal to the square root of eight 𝑥 plus three, which of course means that 𝑧 squared will just be equal to eight 𝑥 plus three. So by using the fact that 𝑧 squared is eight 𝑥 plus three, we get d𝑦 by d𝑥 is equal to 24 divided by three plus nine times eight 𝑥 plus three all squared.

And there’s only a little bit more simplification we need to do. We’ll distribute nine over our parentheses. This gives us 72𝑥 plus 27. And finally we have three plus 27 is equal to 30. And this is our final answer. Therefore, by using the chain rule, the quotient rule, and the general power rule, we were able to show if 𝑦 is equal to 𝑧 squared over three plus nine 𝑧 squared and 𝑧 is equal to the square root of eight 𝑥 plus three, then d𝑦 by d𝑥 is equal to 24 divided by 72𝑥 plus 30 all squared.

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