Video: Finding the Second Derivative of Rational Functions

If 𝑦 : 𝑦 = (βˆ’π‘₯⁡ βˆ’ 1)/(βˆ’π‘₯⁡ + 1), find 𝑦″.

06:13

Video Transcript

If 𝑦 is such that 𝑦 is equal to negative π‘₯ to the fifth power minus one all divided by negative π‘₯ to the fifth power plus one, find 𝑦 double prime.

We’re given 𝑦 as a function in π‘₯. We can see it’s the quotient of two polynomials, so 𝑦 is a rational function in π‘₯. And we’re asked to find 𝑦 double prime. That’s the second derivative of 𝑦 with respect to π‘₯. And since 𝑦 is given as a rational function, we have several different options. For example, we could simplify our expression for 𝑦 by using algebraic division and then use the chain rule or the general power rule. However, in this case, it’s just easier to use the quotient rule.

So we’ll start by recalling the quotient rule. This tells us the derivative of the quotient of two functions 𝑒 of π‘₯ over 𝑣 of π‘₯ with respect to π‘₯ is equal to 𝑒 prime of π‘₯ times 𝑣 of π‘₯ minus 𝑣 prime of π‘₯ times 𝑒 of π‘₯ all divided by 𝑣 of π‘₯ all squared. And we can use this to find the first derivative of 𝑦 with respect to π‘₯. We need to set 𝑒 of π‘₯ to be the function in our numerator and 𝑣 of π‘₯ to be the function in our denominator. So that’s 𝑒 of π‘₯ is negative π‘₯ to the fifth power minus one and 𝑣 of π‘₯ is negative π‘₯ to the fifth power plus one.

And we can see to use the quotient rule, we’re going to need to find expressions for 𝑒 prime of π‘₯ and 𝑣 prime of π‘₯. We can find both of these by using the power rule for differentiation. We want to multiply by our exponent of π‘₯ and then reduce this exponent by one. This gives us that 𝑒 prime of π‘₯ is negative five π‘₯ to the fourth power and 𝑣 prime of π‘₯ is also negative five π‘₯ to the fourth power. Now that we found expressions for 𝑒 prime of π‘₯ and 𝑣 prime of π‘₯, we can find an expression for 𝑦 prime by using the quotient rule. It’s equal to negative five π‘₯ to the fourth power times negative π‘₯ to the fifth power plus one minus negative five π‘₯ to the fourth power multiplied by negative π‘₯ to the fifth power minus one all divided by negative π‘₯ to the fifth power plus one all squared.

And we can simplify this expression. We’ll distribute negative five π‘₯ to the fourth power over both of our sets of parentheses. Distributing over our first set of parentheses, we get five π‘₯ to the ninth power minus five π‘₯ to the fourth power. Distributing over our second set of parentheses and remembering we need to subtract this means we’re subtracting five π‘₯ to the ninth power plus five π‘₯ to the fourth power. And then, of course, we need to divide all of this through by negative π‘₯ to the fifth power plus one all squared. And we can simplify this even further. We’ll distribute the negative one over our second set of parentheses. So we’re now subtracting negative five π‘₯ to the ninth power and negative five π‘₯ to the fourth power in our numerator.

And finally, we can start canceling. First, five π‘₯ to the ninth power minus five π‘₯ to the ninth power is equal to zero. Next, negative five π‘₯ to the fourth power minus negative five π‘₯ to the fourth power is equal to negative 10π‘₯ to the fourth power. So we were able to show that 𝑦 prime is equal to negative 10π‘₯ to the fourth power all divided by negative π‘₯ to the fifth power plus one all squared. But remember, we’re trying to find an expression for 𝑦 double prime. That means we need to differentiate this expression for 𝑦 prime with respect to π‘₯. And since this is the quotient of two polynomials, we’ll do this once again by using the quotient rule.

To do this, we’ll start by calling the numerator of this function 𝑒 two of π‘₯ and the denominator of this function 𝑣 two of π‘₯. So we have 𝑒 two of π‘₯ is negative 10π‘₯ to the fourth power and 𝑣 two of π‘₯ is negative π‘₯ to the fifth power plus one all squared. We’ll also call our original functions 𝑒 of π‘₯ and 𝑣 of π‘₯ 𝑒 one of π‘₯ and 𝑣 one of π‘₯. Now we once again need to apply the quotient rule, so we’re going to need to find expressions for 𝑒 two prime and 𝑣 two prime. We’ll start with 𝑒 two prime. Once again, we can do this by using the power rule for differentiation. We get 𝑒 two prime is equal to negative 40π‘₯ cubed. Now we also need to find an expression for 𝑣 two prime.

There’s several different ways we could do this. For example, we could distribute the square over our parentheses and then use the power rule for differentiation. However, we can also find 𝑣 two prime by using the chain rule or the general power rule. We’ll use the general power rule. We recall this tells us for any real constant 𝑛 and differentiable function 𝑓 of π‘₯, the derivative of 𝑓 of π‘₯ all raised to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times 𝑓 prime of π‘₯ multiplied by 𝑓 of π‘₯ raised to the power of 𝑛 minus one. In our case, our exponent 𝑛 is equal to two and our inner function is negative π‘₯ to the fifth power plus one. We recall this is 𝑣 one of π‘₯.

This means we’ve already found an expression for 𝑓 prime of π‘₯. It’s equal to negative five π‘₯ to the fourth power. So by substituting in our value for 𝑛 and our expressions for 𝑓 prime of π‘₯ and 𝑓 of π‘₯, we get 𝑣 two prime of π‘₯ is equal to two times negative five π‘₯ to the fourth power multiplied by negative π‘₯ to the fifth power plus one all raised to the power of two minus one. And of course, we can simplify this. First, in our exponent, two minus one is equal to one, and raising a function to the first power doesn’t change its value. Next, we have two multiplied by negative five is equal to negative 10. So we’ve shown that 𝑣 two prime of π‘₯ is equal to negative 10π‘₯ to the fourth power multiplied by negative π‘₯ to the fifth power plus one.

Now that we found expressions for 𝑒 two prime and 𝑣 two prime, we’re ready to find an expression for 𝑦 double prime by using the quotient rule. We substitute our expressions for 𝑒 two, 𝑣 two, 𝑒 two prime, and 𝑣 two prime into our quotient rule to give us that 𝑦 double prime is equal to the following expression. And now we can start simplifying. First, in our denominator, we have negative π‘₯ to the fifth power plus one all squared, and then this is all squared. We can just replace this with negative π‘₯ to the fifth power plus one all raised to the fourth power.

Next, in our numerator, we have negative one multiplied by negative 10π‘₯ to the fourth power multiplied by negative 10π‘₯ to the fourth power. This is equal to negative 100π‘₯ to the eighth power. This gives us that 𝑦 double prime is equal to the following expression. And we can now see that our numerator and our denominator share a factor of negative π‘₯ to the fifth power plus one. So we want to cancel one shared factor of negative π‘₯ to the fifth power plus one in our numerator and our denominator. Doing this gives us that 𝑦 double prime is equal to negative 40π‘₯ cubed times negative π‘₯ to the fifth power plus one minus 100π‘₯ to the eighth power all divided by negative π‘₯ to the fifth power plus one all cubed.

And we can simplify this even further. First, we’ll distribute negative 40π‘₯ cubed over our parentheses. And doing this and simplifying, we get 40π‘₯ to the eighth power minus 40π‘₯ cubed. And the last piece of simplification we’ll do is 40π‘₯ to the eighth power minus 100π‘₯ to the eighth power is equal to negative 60π‘₯ to the eighth power. And doing this gives us our final answer. Therefore, by using the quotient rule twice, we were able to show if 𝑦 is equal to negative π‘₯ to the fifth power minus one divided by negative π‘₯ to the fifth power plus one, then 𝑦 double prime will be equal to negative 60π‘₯ to the eighth power minus 40π‘₯ cubed all divided by negative π‘₯ to the fifth power plus one all cubed.

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