Video: Differentiating Rational Functions Using the Quotient Rule

Find 𝑑𝑦/𝑑π‘₯ if 𝑦 = (π‘₯Β² + 3)/(π‘₯Β³ + 3).

02:17

Video Transcript

Find 𝑑𝑦 by 𝑑π‘₯ if 𝑦 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 𝑓 of π‘₯ over 𝑔 of π‘₯ in terms of 𝑓 of π‘₯ 𝑔 of π‘₯ and their derivatives.

Let’s compare 𝑓 of π‘₯ over 𝑔 of π‘₯ to what we have to differentiate to see what 𝑓 of π‘₯ and 𝑔 of π‘₯ are. 𝑓 of π‘₯ is the numerator π‘₯ squared plus three and 𝑔 of π‘₯ is the denominator π‘₯ cubed plus three. To apply the quotient rule, we also need the derivatives of 𝑓 of π‘₯ and 𝑔 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 𝑛, with respect to π‘₯ is 𝑛 times π‘₯ to the 𝑛 minus one and the derivative of a constant function 𝑐 with respect to π‘₯ is zero.

Applying these rules again, we find that the derivative of 𝑔 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 𝑓 of π‘₯ and π‘₯ cubed plus three for 𝑔 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 let’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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.