Question Video: Differentiating Polynomial Functions in the Factored Form | Nagwa Question Video: Differentiating Polynomial Functions in the Factored Form | Nagwa

# Question Video: Differentiating Polynomial Functions in the Factored Form Mathematics • Second Year of Secondary School

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Differentiate π(π₯) = 7π₯Β²(β3π₯ + 9).

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### Video Transcript

Differentiate π of π₯ is equal to seven π₯ squared times negative three π₯ plus nine.

We need to differentiate the function π of π₯. However, we can immediately see a problem. π of π₯ is the product of two functions. Itβs seven π₯ squared multiplied by negative three π₯ plus nine. And we donβt know how to differentiate the product of two functions. So it seems like we canβt answer this question.

However, we do know how to manipulate this expression. We could distribute seven π₯ squared over our parentheses. This would then give us a polynomial. And we do know how to differentiate polynomials. So weβll start by distributing seven π₯ squared over our parentheses. This gives us π of π₯ is equal to seven π₯ squared times negative three π₯ plus seven π₯ squared multiplied by nine. And we can then simplify this expression. Our first term seven π₯ squared multiplied by negative three π₯ is equal to negative 21π₯ cubed.

Remember, this is because when we multiply π₯ squared by π₯, we can just add their exponents together. And we can rewrite π₯ as π₯ to the first power. And we can simplify our second term. Nine times seven is equal to 63. So our second term simplifies to give us 63π₯ squared.

Now, weβve written our function π of π₯ as a polynomial. So we can differentiate this term by term by using the power rule for differentiation. We have the derivative of π with respect to π₯ will be equal to the derivative of negative 21π₯ cubed with respect to π₯ plus the derivative of 63π₯ squared with respect to π₯. And to differentiate these, we need to recall the power rule for differentiation, which tells us for any real constants π and π, the derivative of ππ₯ to the πth power with respect to π₯ is equal to ππ times π₯ to the power of π minus one. We multiply by our exponent of π₯ and then reduce this exponent by one.

In the derivative of our first term, our value of π is equal to three and our value of π is equal to negative 21. So when we differentiate this term, we want to multiply by our exponent of three and then reduce this exponent by one. This gives us three times negative 21π₯ to the power of three minus one. And we can do the same in the derivative of our second term. Our value of π is equal to two, and our value of π is equal to 63.

So once again, when we differentiate this term, we want to multiply by our exponent of two and then reduce this exponent by one. This gives us two times 63 multiplied by π₯ to the power of two minus one. And now we can simplify. Three times negative 21 is equal to negative 63. And three minus one is equal to two. So our first term simplifies to give us negative 63π₯ squared. And in our second term, two times 63 is equal to 126. And two minus one is equal to one. So our second term simplifies to give us 126π₯ to the first power. And π₯ to the first power is just equal to π₯.

Therefore, by distributing seven π₯ squared over our parentheses to rewrite our function π of π₯ as a polynomial, we were able to differentiate π of π₯ is equal to seven π₯ squared times negative three π₯ plus nine. We got that π prime of π₯ is equal to negative 63π₯ squared plus 126π₯.

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