Video: Differentiating Rational Functions Using the Quotient Rule

Differentiate 𝑓(π‘₯) = (5π‘₯Β² βˆ’ 1)/(7π‘₯ + 6).

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Video Transcript

Differentiate the function 𝑓 of π‘₯ which is equal to five π‘₯ squared minus one over seven π‘₯ plus six.

The first thing that stands out to us in this question is that we have the quotient of two differentiable functions, the function five π‘₯ squared minus one in the numerator and the function seven π‘₯ plus six in the denominator. To differentiate functions of this type, we need to use the quotient rule. This tells us that, for two differentiable functions 𝑒 and 𝑣, the derivative of their quotient, 𝑒 over 𝑣, is equal to 𝑣 multiplied by 𝑒 prime minus 𝑒 multiplied by 𝑣 prime all over 𝑣 squared.

Let’s see how to apply the quotient rule in this question. Remember, 𝑒 is the function in the numerator of the quotient and 𝑣 is the function in the denominator. So we’re going to let 𝑒 equal the function in the numerator, five π‘₯ squared minus one, and 𝑣 equal the function in the denominator, seven π‘₯ plus six. We then need to find each of their individual derivatives with respect to π‘₯, which we can do using the power rule of differentiation. 𝑒 prime or d𝑒 by dπ‘₯ is equal to five multiplied by two π‘₯, which is 10π‘₯. And remember, the derivative of a constant, in this case negative one, is just zero. 𝑣 prime or d𝑣 by dπ‘₯ is equal to seven. So we have both the derivatives of 𝑒 and 𝑣 with respect to π‘₯.

We’re now ready to substitute into the quotient rule to find 𝑓 prime of π‘₯. It’s equal to 𝑣 multiplied by 𝑒 prime. That’s seven π‘₯ plus six multiplied by 10π‘₯ minus 𝑒 multiplied by 𝑣 prime. That’s five π‘₯ squared minus one multiplied by seven. And this is all divided by 𝑣 squared. That’s seven π‘₯ plus six all squared. We now just need to distribute the parentheses in the numerator and simplify the result. We obtain 70π‘₯ squared plus 60π‘₯ minus 35π‘₯ squared plus seven all over seven π‘₯ plus six all squared. And then, collecting like terms in the numerator, we have 35π‘₯ squared plus 60π‘₯ plus seven all over seven π‘₯ plus six all squared.

By applying the quotient rule then, we’ve differentiated our function 𝑓 of π‘₯. Notice that it would also be possible to answer this question using the product rule. If we were to express our function 𝑓 of π‘₯ as five π‘₯ squared minus one multiplied by seven π‘₯ plus six to the power of negative one. And if you’re interested, you could answer the question using the product rule and confirm it does indeed give the same result.

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