Question Video: Using the Product Rule | Nagwa Question Video: Using the Product Rule | Nagwa

# Question Video: Using the Product Rule Mathematics • Second Year of Secondary School

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Suppose that π is differentiable. What is the derivative of π₯Β³π(π₯)?

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### Video Transcript

Suppose that π is differentiable. What is the derivative of π₯ cubed π of π₯?

So if we take a look at π₯ cubed π of π₯, what this is is π₯ cubed multiplied by a function. So how are we going to differentiate this? Well to differentiate this, what weβre gonna use is something called the product rule. And weβre gonna use that because the product rule tells us that if we have π¦ is equal to π’π£, so if weβre looking at two things multiplied together like here we have π₯ cubed and π of π₯, then the derivative or ππ¦ ππ₯ is gonna be equal to π’ ππ£ ππ₯ plus π£ ππ’ ππ₯. So ππ’ multiplied by the derivative of π£ plus π£ multiplied by the derivative of π’.

So in our expression, what we have is π₯ cubed, this is gonna be our π’, and π of π₯, this is gonna be our π£. Therefore, if π’ was equal to π₯ cubed, ππ’ ππ₯ is gonna be equal to three π₯ squared. And just to remind us how we got that, we got the exponent, which is three, and then we multiply it by the coefficient, which is one, so three multiplied by one. And then weβve got π₯ to the power of, and then you subtract one from the exponent. So three minus one gives us two, so it gives us our three π₯ squared. And then if π£ is our function of π₯, then ππ£ ππ₯ is gonna be the derivative of this function of π₯, which Iβve shown using π prime of π₯.

So this is gonna give us the derivative is equal to π₯ cubed π prime of π₯ or the derivative of π of π₯. And thatβs because that was our π’ multiplied by our ππ£ ππ₯ plus π£ ππ’ ππ₯, so three π₯ squared multiplied by π of π₯. And thatβs because our π£ was π of π₯ and our ππ’ ππ₯ was three π₯ squared. And therefore, we can write this the other way round if we want to have it in increasing powers of π₯. But we can suppose that if π is differentiable, the derivative of π₯ cubed π of π₯ is going to be three π₯ squared π of π₯ plus π₯ cubed of the derivative of π of π₯.

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