Video: Differentiating Rational Functions Using the Quotient Rule

Find the first derivative of 𝑦 = (8π‘₯ + 5)/(3π‘₯ + 22).

02:33

Video Transcript

Find the first derivative of 𝑦 is equal to 8π‘₯ plus five over three π‘₯ plus 22.

Here, we can see that our function 𝑦 is a rational function. So we can find its derivative by using the quotient rule. The quotient rule tells us that if we differentiate the quotient of two functions, so 𝑒 over 𝑣, with respect to π‘₯. Then it’s equal to 𝑣 multiplied by the differential of 𝑒 with respect to π‘₯ minus 𝑒 timesed by the differential of 𝑣 with respect to π‘₯ all over 𝑣 squared. In order to find the first derivative of 𝑦, let’s start by labeling 𝑒 and 𝑣 from our equation. 𝑒 will be equal to the numerator of the function, so eight π‘₯ plus five. And 𝑣 will be equal to the denominator of the function, so that’s three π‘₯ plus 22.

Next, we must find d by dπ‘₯ of 𝑒 and d by dπ‘₯ of 𝑣, or d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯. 𝑒 and 𝑣 are both polynomials. So we can simply differentiate them term by term. Writing 𝑒 in terms of powers of π‘₯, we can say that it’s equal to eight π‘₯ to the power of one plus five π‘₯ to the power of zero. In order to differentiate, we simply multiply by the power and decrease the power by one. For the first time, we multiply by the power, so that’s one, and decrease the power by one, to zero. Leaving us with one timesed by eight π‘₯ to the power of zero. For the second term, we multiply by the power, so that’s zero, and decrease the power by one, to negative one. Giving us zero multiplied by five π‘₯ to the negative one.

In the first term, π‘₯ to the power of zero is just one. So this becomes eight. In the second term, we’re multiplying by zero. So that term becomes zero. Therefore, we find that d𝑒 by dπ‘₯ is equal to eight. We can use a similar method to find d𝑣 by dπ‘₯. And we find that it’s equal to three. Now that we found d𝑒 by dπ‘₯ and d𝑣 by dπ‘₯, we’re ready to use the quotient rule. We find that d𝑦 by dπ‘₯ is equal to 𝑣, which is three π‘₯ plus 22, multiplied by d𝑒 by dπ‘₯, so that’s eight, minus 𝑒, so that’s eight π‘₯ plus five, multiplied by d𝑣 by dπ‘₯, so that’s three. And this is all over 𝑣 squared, so that’s three π‘₯ plus 22 all squared.

Next, we can expand the brackets. And then simplify to find that our solution is that the first derivative of 𝑦 is equal to 161 over three π‘₯ plus 22 all squared.

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