Video Transcript
Find the second derivative of the function 𝑦 is equal to five 𝑥 minus four all divided by two 𝑥 minus three at the point two, six.
We’re given 𝑦 as a function in 𝑥, and we’re asked to find the second derivative of 𝑦 at the point two, six. Since 𝑦 is given as a function in 𝑥, this means we need to find the second derivative of 𝑦 with respect to 𝑥. In other words, we need to differentiate this twice. And then we need to determine the value of this at the point two, six. So that’s when 𝑥 is equal to two. So we need to differentiate 𝑦 with respect to 𝑥 twice. We can see 𝑦 is a rational function. There’s a couple of different ways we could differentiate this. For example, we could use algebraic division and then the general power rule. However, we’re going to do this by using the quotient rule.
So let’s start by recalling the quotient rule for differentiation. If we have two differentiable functions 𝑢 of 𝑥 and 𝑣 of 𝑥, then the derivative of 𝑢 of 𝑥 divided by 𝑣 of 𝑥 with respect to 𝑥 is equal to 𝑢 prime of 𝑥 times 𝑣 of 𝑥 minus 𝑣 prime of 𝑥 times 𝑢 of 𝑥 all divided by 𝑣 of 𝑥 all squared. We want to use this to differentiate our function 𝑦. So we need to set 𝑢 to be the function in our numerator and 𝑣 to be the function in our denominator. So that’s 𝑢 of 𝑥 is five 𝑥 minus four and 𝑣 of 𝑥 is two 𝑥 minus three.
Now we can see to apply the quotient rule, we need to find expressions for 𝑢 prime of 𝑥 and 𝑣 prime of 𝑥. And in fact, both 𝑢 of 𝑥 and 𝑣 of 𝑥 are linear functions, so their derivatives with respect to 𝑥 will be the coefficients of 𝑥. So 𝑢 prime of 𝑥 is five, and 𝑣 prime of 𝑥 is two. Now that we found 𝑢 prime of 𝑥 and 𝑣 prime of 𝑥, we can use the quotient rule to find an expression for 𝑦 prime. We just need to substitute in our expressions for 𝑢, 𝑣, 𝑢 prime, and 𝑣 prime into our quotient rule. This gives us five times two 𝑥 minus three minus two times five 𝑥 minus four all over two 𝑥 minus three all squared.
And we can simplify this expression. We’ll start by distributing five over our first set of parentheses and two over our second set of parentheses. Doing this gives us a new numerator of 10𝑥 minus 15 minus 10𝑥 plus eight. And we can simplify this; first, 10𝑥 minus 10𝑥 is equal to zero. Next, negative 15 plus eight is equal the negative seven. So we’ve shown 𝑦 prime is equal to negative seven divided by two 𝑥 minus three all squared. But remember, we need to find the second derivative of 𝑦, so we need to differentiate this one more time.
By differentiating our expression for 𝑦 with respect to 𝑥, we have 𝑦 double prime will be equal to the derivative of negative seven divided by two 𝑥 minus three all squared with respect to 𝑥. And once again, we have a few different options of how we can differentiate this. For example, we could use the quotient rule again, or we could use the chain rule. However, we’re going to use the general power rule. The first step in using the general power rule will be to use our laws of exponents to rewrite our expression. We want to write this as the derivative of negative seven times two 𝑥 minus three all raised to the power of negative two with respect to 𝑥.
Now, all we need to do is recall the general power rule. This tells us for any real constant 𝑛 and differentiable function 𝑓 of 𝑥, the derivative of 𝑓 of 𝑥 all raised to the 𝑛th power with respect to 𝑥 is equal to 𝑛 times 𝑓 prime of 𝑥 multiplied by 𝑓 of 𝑥 raised to the power of 𝑛 minus one. To apply this to our derivative, we need to set the value of our exponent 𝑛 equal to negative two. And our function 𝑓 of 𝑥 will be the inner function two 𝑥 minus three, which we’ve already called 𝑣 of 𝑥. Before we use the general power rule, we can simplify our expression slightly. We can take the constant factor of negative seven outside of our derivative. This gives us the following expression, and we already found an expression for 𝑓 prime of 𝑥. We showed this was equal to two.
Substituting 𝑛 is equal to negative two, 𝑓 of 𝑥 is equal to two 𝑥 minus three, and 𝑓 prime of 𝑥 is equal to two into our general power rule, we get that 𝑦 double prime is equal to negative seven times negative two multiplied by two times two 𝑥 minus three all raised to the power of negative two minus one. And we can simplify this expression slightly. It’s equal to 28 times two 𝑥 minus three all raised to the power of negative three. But remember the question wants to us evaluate the second derivative at the point two, six. So we need to substitute 𝑥 is equal to two into this expression.
Doing this gives us 𝑦 double prime of two is equal to 28 times two times two minus three all raised to the power of negative three. And we can simplify this expression. First, two times two minus three is equal to one. So this simplifies to give us 28 times one to the power of negative three, and one raised to any power is just equal to one. So this simplifies to give us 28. Therefore, we were able to show if 𝑦 is equal to five 𝑥 minus four all divided by two 𝑥 minus three, then the second derivative of 𝑦 with respect to 𝑥 at the point two, six is equal to 28.
