Question Video: Finding the Second Derivative of Rational Functions | Nagwa Question Video: Finding the Second Derivative of Rational Functions | Nagwa

# Question Video: Finding the Second Derivative of Rational Functions Mathematics • Third Year of Secondary School

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Given that π¦ = (3π₯Β² β 5)/(2π₯Β² + 7), determine dΒ²π¦/dπ₯Β².

03:18

### Video Transcript

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

Here we have a quotient. Itβ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 π’ and π£, the derivative of π’ over π£ with respect to π₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared.

Since π’ is the numerator, weβre going to let π’ be equal to three π₯ squared minus five. And we can see that π£ must be equal to two π₯ squared plus seven. To be able to use the quotient rule, weβre going to differentiate each of these with respect to π₯. When we differentiate π’ with respect to π₯, we get six π₯. And dπ£ by dπ₯ is equal to four π₯. π£ times dπ’ by dπ₯ is two π₯ squared plus seven times six π₯. We then subtract the product of π’ and dπ£ by dπ₯. Thatβs three π₯ squared minus five times four π₯. And, of course, thatβs all over π£ squared. Thatβ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, weβll keep the denominator as is. And we see that the derivative of π¦ with respect π₯ is 62π₯ over two π₯ squared plus seven squared.

To find the second derivative, weβre going to need to differentiate this again. Letβs clear some space. Once again, weβre looking to differentiate a quotient, so weβre going to use the quotient rule. This time, we let π’ be equal to 62π₯ and π£ be equal to two π₯ squared plus seven squared. dπ’ by dπ₯ is fairly straightforward. Itβs 62. But what about dπ£ by dπ₯?

Well, we can use a special case of the chain rule called the general power rule. This says that if π is some function of π₯ and π is some constant rational not equal to zero, then the derivative of π to the power of π with respect to π₯ is π times π to the power of π minus one times dπ by dπ₯. This means that dπ£ 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 dπ£ by dπ₯ is eight π₯ times two π₯ squared plus seven. This time, π£ times dπ’ by dπ₯ is two π₯ squared plus seven squared times 62 minus π’ times dπ£ by dπ₯ all over π£ 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 π¦ with respect to π₯ is 62 times seven minus six squared over two π₯ squared plus seven cubed.

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