# Question Video: Finding the Second Derivative of Rational Functions

Given that 𝑦 = (3𝑥² − 5)/(2𝑥² + 7), determine d²𝑦/d𝑥².

03:18

### Video Transcript

Given that 𝑦 is equal to three 𝑥 squared minus five over two 𝑥 squared plus seven, determine the second derivative of 𝑦 with respect to 𝑥.

Here we have a quotient. It’s the result of dividing one function by another function. We can, therefore, use the quotient rule to help us find the first derivative. This says that for two differentiable functions 𝑢 and 𝑣, the derivative of 𝑢 over 𝑣 with respect to 𝑥 is equal to 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 squared.

Since 𝑢 is the numerator, we’re going to let 𝑢 be equal to three 𝑥 squared minus five. And we can see that 𝑣 must be equal to two 𝑥 squared plus seven. To be able to use the quotient rule, we’re going to differentiate each of these with respect to 𝑥. When we differentiate 𝑢 with respect to 𝑥, we get six 𝑥. And d𝑣 by d𝑥 is equal to four 𝑥. 𝑣 times d𝑢 by d𝑥 is two 𝑥 squared plus seven times six 𝑥. We then subtract the product of 𝑢 and d𝑣 by d𝑥. That’s three 𝑥 squared minus five times four 𝑥. And, of course, that’s all over 𝑣 squared. That’s two 𝑥 squared plus seven squared.

Distributing the parentheses on the top of our fraction, and we get 12𝑥 cubed plus 42𝑥 minus 12𝑥 cubed plus 20𝑥. And then, we’ll keep the denominator as is. And we see that the derivative of 𝑦 with respect 𝑥 is 62𝑥 over two 𝑥 squared plus seven squared.

To find the second derivative, we’re going to need to differentiate this again. Let’s clear some space. Once again, we’re looking to differentiate a quotient, so we’re going to use the quotient rule. This time, we let 𝑢 be equal to 62𝑥 and 𝑣 be equal to two 𝑥 squared plus seven squared. d𝑢 by d𝑥 is fairly straightforward. It’s 62. But what about d𝑣 by d𝑥?

Well, we can use a special case of the chain rule called the general power rule. This says that if 𝑝 is some function of 𝑥 and 𝑛 is some constant rational not equal to zero, then the derivative of 𝑝 to the power of 𝑛 with respect to 𝑥 is 𝑛 times 𝑝 to the power of 𝑛 minus one times d𝑝 by d𝑥. This means that d𝑣 by d𝑥 is two times this function in 𝑥, which is two 𝑥 squared plus seven to the power of one, times the derivative of two 𝑥 squared plus seven with respect 𝑥, which is four 𝑥.

And when we simplify this, we see that d𝑣 by d𝑥 is eight 𝑥 times two 𝑥 squared plus seven. This time, 𝑣 times d𝑢 by d𝑥 is two 𝑥 squared plus seven squared times 62 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 squared. The key here is to spot that the denominator of our fraction becomes two 𝑥 squared plus seven to the power of four. And this means we can divide through by a common factor of two 𝑥 squared plus seven. And that leaves us with two 𝑥 squared plus seven cubed on the denominator and 62 times two 𝑥 squared plus seven minus 62𝑥 times eight 𝑥 on the top.

We can take out a factor of 62 on the numerator. And then, when we subtract eight 𝑥 times 𝑥 from two 𝑥 squared, we get negative six 𝑥 squared. And the second derivative of 𝑦 with respect to 𝑥 is 62 times seven minus six squared over two 𝑥 squared plus seven cubed.