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

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

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Given that π¦ =β(4π₯Β² β 5) and π§ = 5π₯Β² + 9, determine (π¦(dπ¦/dπ₯)) + (dπ§/dπ₯).

03:45

### Video Transcript

Given that π¦ is equal to the square root of four π₯ squared minus five and π§ is equal to five π₯ squared plus nine, determine π¦ multiplied by the derivative of π¦ with respect to π₯ plus the derivative of π§ with respect to π₯

The question gives us an equation for π¦ in terms of π₯ and an equation for π§ in terms of π₯. And the expression we are asked to evaluate only has derivatives with respect to π₯, 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 π raised to the πth power is equal to π multiplied by π prime of π₯ multiplied by π of π₯ raised to the power of π minus one. And second, we have a particular case of this rule. The derivative of π₯ raised to the πth power is π multiplied by π₯ to the power of π minus one.

So, letβs start by calculating the derivative of π¦ with respect to π₯. We have that this is equal to the derivative of the square root of four π₯ squared minus five with respect to π₯. And we can notice that taking a square root is equivalent to raising something to the power of a half. What this means is weβre differentiating the function of the form π of π₯ raised to the πth power, where π is equal to a half, and π of π₯ is equal to four π₯ squared minus five.

So, to differentiate this, we use our derivative rule. First, we multiply it by π, which is a half. Next, we multiply this by the derivative of our function π of π₯, which is four π₯ squared minus five. And finally, we multiply this by π of π₯ raised to the power of π minus one. So, we get four π₯ 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 π₯ squared minus five with respect to π₯, which we can calculate by using our derivative rule to just be eight π₯.

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 π₯ multiplied by four π₯ squared minus five raised to the power of negative a half. Next, we can simplify a half multiplied by eight π₯ to be four π₯. And instead of multiplying by four π₯ 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 π¦ with respect to π₯ is equal to four π₯ divided by four π₯ squared minus five raised to the power of a half.

Now, letβs calculate the derivative of π§ with respect to π₯. So, thatβs the derivative of five π₯ squared plus nine with respect to π₯. We can calculate this derivative directly using our power rule for differentiation. This gives us that derivative of π§ with respect to π₯ is equal to 10π₯. We are now ready to evaluate the expression given to us in the question. That is π¦ multiplied by the derivative of π¦ with respect to π₯ plus the derivative of π§ with respect to π₯.

We start by substituting in that π¦ is equal to the square root of four π₯ squared minus five. Next, we substitute in the expression we calculated for the derivative of π¦ with respect to π₯. So, thatβs four π₯ divided by four π₯ squared minus five raised to the power of a half. And then, we substitute in our expression for dπ§ by dπ₯, which in this case is 10π₯. We can then notice that our denominator, four π₯ squared minus five all raised to the power of a half, is actually equal to the square root of four π₯ squared minus five.

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

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