Question Video: Differentiating Rational Functions Using the Quotient Rule | Nagwa Question Video: Differentiating Rational Functions Using the Quotient Rule | Nagwa

# Question Video: Differentiating Rational Functions Using the Quotient Rule Mathematics • Second Year of Secondary School

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Find ππ¦/ππ₯ if π¦ = (π₯Β² + 3)/(π₯Β³ + 3).

02:17

### Video Transcript

Find ππ¦ by ππ₯ if π¦ is equal to π₯ squared plus three over π₯ cubed plus three.

Here we have to differentiate a rational function and so we need to use the quotient rule for differentiation which allows us to write the derivative of a quotient of two functions π of π₯ over π of π₯ in terms of π of π₯ π of π₯ and their derivatives.

Letβs compare π of π₯ over π of π₯ to what we have to differentiate to see what π of π₯ and π of π₯ are. π of π₯ is the numerator π₯ squared plus three and π of π₯ is the denominator π₯ cubed plus three. To apply the quotient rule, we also need the derivatives of π of π₯ and π of π₯. The derivative of π₯ squared plus three is two π₯, which we can get by using the fact that the derivative of a power of π₯, π₯ to the π, with respect to π₯ is π times π₯ to the π minus one and the derivative of a constant function π with respect to π₯ is zero.

Applying these rules again, we find that the derivative of π of π₯ with respect to π₯ is three π₯ squared. Now, we have all the ingredients needed to apply the quotient rule. We substitute π₯ squared plus three for π of π₯ and π₯ cubed plus three for π of π₯ on the left-hand side. Continuing to substitute, the first term in the numerator becomes π₯ cubed plus three times two π₯. And from this, we subtract π₯ squared plus three times three π₯ squared and we divide this by π₯ cubed plus three squared.

Now letβs expand and simplify in the numerator. Expanding, we get two π₯ to the four plus six π₯ minus three π₯ to the four minus nine π₯ squared. And we see that there are some like terms that we can combine to get our final answer minus π₯ to the four minus nine π₯ squared plus six π₯ all over π₯ cubed plus three squared.

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