Video: Differentiating Rational Functions Using the Quotient Rule

Differentiate 𝑦 = (2π‘₯⁷ βˆ’ 6π‘₯⁡ + 4π‘₯Β³ βˆ’ 8)/4π‘₯⁡.

05:43

Video Transcript

Differentiate 𝑦 is equal to two π‘₯ to the seventh power minus six π‘₯ to the fifth power plus four π‘₯ cubed minus eight all divided by four π‘₯ to the fifth power.

The question wants us to find the derivative of the quotient of two polynomials. And since the question wants us to differentiate the quotient of two functions, we’ll do this by using the quotient rule. We recall the quotient rule tells us the derivative of the quotient of two functions 𝑒 over 𝑣 is equal to 𝑣 times 𝑒 prime minus 𝑒 times 𝑣 prime all divided by 𝑣 squared.

So, to apply the quotient rule, we’ll set 𝑒 of π‘₯ to be the polynomial in our numerator. That’s two π‘₯ to the seventh power minus six π‘₯ to the fifth power plus four π‘₯ cubed minus eight. And we’ll set 𝑣 of π‘₯ to be the polynomial in our denominator. That’s four π‘₯ to the fifth power. Now, to apply the quotient rule, we’re going to need to find expressions for 𝑒 prime of π‘₯ and 𝑣 prime of π‘₯. Since both of these are polynomials, we can do this term by term by using the power rule for differentiation.

We recall this tells us for constants π‘Ž and 𝑛, the derivative of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times π‘Ž times π‘₯ to the power of 𝑛 minus one. We multiply by the exponent of π‘₯ and then reduce this exponent by one. Applying this rule to each term in 𝑒 of π‘₯, we get 𝑒 prime of π‘₯ is equal to 14π‘₯ to the sixth power minus 30π‘₯ to the fourth power plus 12π‘₯ squared. And we can do the same to find an expression for 𝑣 prime of π‘₯, we get 20π‘₯ to the fourth power.

We’re now ready to find an expression for the derivative of 𝑦 with respect to π‘₯ by using the quotient rule. We get that it’s equal to 𝑣 of π‘₯ times 𝑒 prime of π‘₯ minus 𝑣 prime of π‘₯ times 𝑒 of π‘₯ all divided by 𝑣 of π‘₯ squared. Substituting in our expressions for 𝑣 of π‘₯, 𝑒 of π‘₯, 𝑒 prime of π‘₯, and 𝑣 prime of π‘₯. We get that 𝑦 prime of π‘₯ is equal to four π‘₯ to the fifth power times 14π‘₯ to the sixth power minus 30π‘₯ to the fourth power plus 12π‘₯ squared minus 20π‘₯ to the fourth power. Multiplied by two π‘₯ to the seventh power minus six π‘₯ to the fifth power plus four π‘₯ cubed minus eight all divided by four π‘₯ to the fifth power all squared.

It might be tempting to start distributing over our parentheses. However, it’s easier to simplify this expression by factoring. Initially, we’ll simplify our denominator. We have four π‘₯ to the fifth power all squared is equal to 16π‘₯ to the power of 10. We could now start simplifying. First, our numerator and our denominator share a factor of four π‘₯ to the fourth power. So, let’s divide our numerator and our denominator by four π‘₯ to the fourth power.

First, dividing the first term in our numerator by four π‘₯ to the fourth power, we’re left with π‘₯. Next, to divide the second term in our numerator by four π‘₯ to the fourth power, we’ll simplify the first factor to just be five. Finally, we need to divide our denominator by four π‘₯ to the fourth power. We see this leaves us with four π‘₯ to the sixth power, and this gives us the following expression. At this point, there’s several different methods we could use to simplify this expression even further. We could continue factoring our numerator; however, we’ll simplify this by distributing over our parentheses and our numerator.

So, let’s start by distributing π‘₯ over the parentheses in our first term. We need to multiply each term inside of our parentheses by π‘₯. And to do this, we just need to add one to each of our exponents of π‘₯. This gives us 14π‘₯ to the seventh power minus 30π‘₯ to the fifth power plus 12π‘₯ cubed. We now want to do the same with the second term in our numerator. However, remember, this time, there are four terms inside of our parentheses. And we need to multiply each term by negative five. Doing this and being careful with our signs, we get negative 10π‘₯ to the seventh power plus 30π‘₯ to the fifth power minus 20π‘₯ cubed plus 40.

Finally, remember, we need to divide all of this by four π‘₯ to the sixth power. We can now simplify our numerator by combining like terms. First, negative 30π‘₯ to the fifth power plus 30π‘₯ to the fifth power is equal to zero. Next, if we have 14π‘₯ to the seventh power and we subtract 10π‘₯ to the seventh power, we’re just left with four π‘₯ to the seventh power. Similarly, if we have 12π‘₯ cubed and then we subtract 20π‘₯ cubed, we’ll be left with negative eight π‘₯ cubed. So, we’ve now simplified this expression to give us four π‘₯ to the seventh power minus eight π‘₯ cubed plus 40 all divided by four π‘₯ to the sixth power.

And we could leave our answer like this. However, we’ll divide each term in our numerator through by our denominator. So, let’s do this term by term. First, to divide four π‘₯ to the seventh power by four π‘₯ to the sixth power, we can cancel the shared factor of four and we can cancel six of the shared factors of π‘₯. This just leaves us with π‘₯. Next, in our second term, we have eight divided by four will leave us with two and then π‘₯ cubed divided by π‘₯ to the sixth power will leave us with π‘₯ cubed in our denominator. So, our second term simplify to give us negative two divided by π‘₯ cubed.

Finally, in our third term, we have 40 divided by four is equal to 10. So, our third term simplifies to give us 10 divided by π‘₯ to the sixth power. And again, we could leave our answer like this. However, we’ll use our laws of exponents to rewrite our divisions as multiplications. So, by using the fact that dividing by π‘₯ cubed is the same as multiplying by π‘₯ to the power of negative three and dividing by π‘₯ to the sixth power is the same as multiplying by π‘₯ to the power of negative six. We were able to rewrite our answer as π‘₯ minus two π‘₯ to the power of negative three plus 10π‘₯ to the power of negative six. And this is our final answer.

Therefore, we were able to show by using the quotient rule. If 𝑦 is equal to two π‘₯ to the seventh power minus six π‘₯ to the fifth power plus four π‘₯ cubed minus eight all divided by four π‘₯ to the fifth power. Then the derivative of 𝑦 with respect to π‘₯ is equal to π‘₯ minus two π‘₯ to the power of negative three plus 10π‘₯ to the power of negative six.

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