Question Video: Evaluating an Expression Containing Derivatives of a Root and Polynomial Functions Using the Chain Rule | Nagwa Question Video: Evaluating an Expression Containing Derivatives of a Root and Polynomial Functions Using the Chain Rule | Nagwa

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Question Video: Evaluating an Expression Containing Derivatives of a Root and Polynomial Functions Using the Chain Rule Mathematics • Second Year of Secondary School

Given that 𝑦 =√(4𝑥² − 5) and 𝑧 = 5𝑥² + 9, determine (𝑦(d𝑦/d𝑥)) + (d𝑧/d𝑥).

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Video Transcript

Given that 𝑦 is equal to the square root of four 𝑥 squared minus five and 𝑧 is equal to five 𝑥 squared plus nine, determine 𝑦 multiplied by the derivative of 𝑦 with respect to 𝑥 plus the derivative of 𝑧 with respect to 𝑥

The question gives us an equation for 𝑦 in terms of 𝑥 and an equation for 𝑧 in terms of 𝑥. And the expression we are asked to evaluate only has derivatives with respect to 𝑥, so we can attempt to do this directly. To help us calculate this, we recall some derivative rules. First, by using the chain rule we have that the derivative of a function 𝑓 raised to the 𝑛th power is equal to 𝑛 multiplied by 𝑓 prime of 𝑥 multiplied by 𝑓 of 𝑥 raised to the power of 𝑛 minus one. And second, we have a particular case of this rule. The derivative of 𝑥 raised to the 𝑛th power is 𝑛 multiplied by 𝑥 to the power of 𝑛 minus one.

So, let’s start by calculating the derivative of 𝑦 with respect to 𝑥. We have that this is equal to the derivative of the square root of four 𝑥 squared minus five with respect to 𝑥. And we can notice that taking a square root is equivalent to raising something to the power of a half. What this means is we’re differentiating the function of the form 𝑓 of 𝑥 raised to the 𝑛th power, where 𝑛 is equal to a half, and 𝑓 of 𝑥 is equal to four 𝑥 squared minus five.

So, to differentiate this, we use our derivative rule. First, we multiply it by 𝑛, which is a half. Next, we multiply this by the derivative of our function 𝑓 of 𝑥, which is four 𝑥 squared minus five. And finally, we multiply this by 𝑓 of 𝑥 raised to the power of 𝑛 minus one. So, we get four 𝑥 squared minus five all raised to the power of a half minus one. We can simplify this expression by first calculating the derivative of four 𝑥 squared minus five with respect to 𝑥, which we can calculate by using our derivative rule to just be eight 𝑥.

And we can also simplify our exponent, since a half minus one is just equal to negative one-half. This gives us a half multiplied by eight 𝑥 multiplied by four 𝑥 squared minus five raised to the power of negative a half. Next, we can simplify a half multiplied by eight 𝑥 to be four 𝑥. And instead of multiplying by four 𝑥 squared minus five raised to the power of negative a half, we can divide by this raised to the power of positive one-half. This gives us that the derivative of 𝑦 with respect to 𝑥 is equal to four 𝑥 divided by four 𝑥 squared minus five raised to the power of a half.

Now, let’s calculate the derivative of 𝑧 with respect to 𝑥. So, that’s the derivative of five 𝑥 squared plus nine with respect to 𝑥. We can calculate this derivative directly using our power rule for differentiation. This gives us that derivative of 𝑧 with respect to 𝑥 is equal to 10𝑥. We are now ready to evaluate the expression given to us in the question. That is 𝑦 multiplied by the derivative of 𝑦 with respect to 𝑥 plus the derivative of 𝑧 with respect to 𝑥.

We start by substituting in that 𝑦 is equal to the square root of four 𝑥 squared minus five. Next, we substitute in the expression we calculated for the derivative of 𝑦 with respect to 𝑥. So, that’s four 𝑥 divided by four 𝑥 squared minus five raised to the power of a half. And then, we substitute in our expression for d𝑧 by d𝑥, which in this case is 10𝑥. We can then notice that our denominator, four 𝑥 squared minus five all raised to the power of a half, is actually equal to the square root of four 𝑥 squared minus five.

So, we can cancel this shared factor in our numerator and our denominator, giving us four 𝑥 plus 10𝑥, which we can evaluate to give us 14𝑥. Therefore, we have shown that if 𝑦 is equal to the square root of four squared minus five, and 𝑧 is equal to five 𝑥 squared plus nine. Then the expression 𝑦 multiplied by the derivative of 𝑦 with respect to 𝑥 plus the derivative of 𝑧 with respect to 𝑥 is actually equal to 14𝑥.

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