Question Video: Finding the Second Derivative of Rational Functions

Given that 𝑦 = (3π‘₯Β² βˆ’ 5)/(2π‘₯Β² + 7), determine d²𝑦/dπ‘₯Β².

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Video Transcript

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

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 𝑒 and 𝑣, the derivative of 𝑒 over 𝑣 with respect to π‘₯ is equal to 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared.

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

Distributing the parentheses on the top of our fraction, and we get 12π‘₯ cubed plus 42π‘₯ minus 12π‘₯ cubed plus 20π‘₯. And then, we’ll keep the denominator as is. And we see that the derivative of 𝑦 with respect π‘₯ is 62π‘₯ over two π‘₯ 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 𝑒 be equal to 62π‘₯ and 𝑣 be equal to two π‘₯ squared plus seven squared. d𝑒 by dπ‘₯ is fairly straightforward. It’s 62. But what about d𝑣 by dπ‘₯?

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 π‘₯ and 𝑛 is some constant rational not equal to zero, then the derivative of 𝑝 to the power of 𝑛 with respect to π‘₯ is 𝑛 times 𝑝 to the power of 𝑛 minus one times d𝑝 by dπ‘₯. This means that d𝑣 by dπ‘₯ is two times this function in π‘₯, which is two π‘₯ squared plus seven to the power of one, times the derivative of two π‘₯ squared plus seven with respect π‘₯, which is four π‘₯.

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

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

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