Second Derivatives

Find y'' ...

Question

Second Derivatives

Find y'' in Exercises 59–64.

y = x(2x + 1)⁴

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the second derivative of Y is equal to 2 X multiplied by the quantity of 5 X plus 3 in quantity cubed. So this problem we need to find the second derivative of Y. And so we're going to start off by calculating the first derivative and then find, use that to find the second derivative. So we're going to start off by calculating Y, and this is going to be equal to the derivative with respect to X of the quantity of 2 X multiplied by the quantity of 5 X plus 3 in quantity, cubed in quantity. So here I'm taking the derivative of a product. So I want to use our product rule. So here we call that for the product rule that if you have function Y that has the form, Y is equal to F of X, multiplied by G of X. Then its derivative Y is equal to F of X, multiplied by G of X. Plus F of X. Multiplied by G of X. So, in our case here, F of X is going to be equal to 2 X and G of X. Is going to be equal to the quantity of 5 X plus 3 in quantity cubed. So we need to calculate F of X, and that's going to be equal to the derivative with respect to X of the quantity of 2 X. And to take this derivative, we'll need to use our constant multiple and power rule. So if we call your constant multiple in power rule, so this is going to be equal to, we'll have our constant, which is 2, multiplied by the exponent of our X variable, which in this case is 1. Then that gets multiplied by X, raised to the quantity of the exponent, which is 1 minus 1 in quantity. And when we simplify this expression, this is just going to be equal to 2. Now we need to calculate G of X, and that's going to be equal to the derivative with respect to X. Of the quantity of 5 X plus 3 in quantity cubed. And so here to take this derivative, we'll need to apply our power rule, our chain rule, or constant multiple rule, as well as our constant rule for derivatives. So doing that, this is going to be equal to. First we'll apply our power rule, so we'll have our exponent 3, and that gets multiplied by the quantity of 5 X plus 3 in quantity, and that gets raised to the quantity of 3 minus 1 in quantity. But now we need to apply our chain rule and multiply this by the derivative with respect to X of the quantity in the parentheses, which is the quantity of 5 X plus 3 in quantity. So this is going to be equal to out in front of the derivative. We're going to have 3 multiplied by the quantity of 5 X plus 3 in quantity squared, and this gets multiplied by the derivative with respect to X of the quantity of 5 X plus 3. So when I take the derivative of 5 X with respect to X, I'm going to need to use our constant multiple and power rule. So doing so, I'll have a constant 5 multiplied by the exponent of the X variable, which is 1, and then that gets multiplied by X raised to the quantity of 1 minus 1. And then we'll have to add to this. The derivative of 3 with respect to X, and here I'm taking a derivative of a constant, and anytime I take a derivative of a constant, I get 0. So this will have plus 0. And so finally simplifying this expression, this is going to be equal to 15 multiplied by the quantity of 5 X plus 3 in quantity squared. So now we can actually calculate our first derivative. So we'll have Y is equal to, and here we'll apply our product rule formula. So we'll have F of X, which is 2, that gets multiplied by G of X, which is the quantity of 5 X plus 3 in quantity cubed. Then we'll have plus F of X, which is 2 X, multiplied by. The quantity of G of X, which is 15, multiplied by the quantity of 5 X plus 3 in quantity squared. So now we just need to simplify this expression. So we can factor out a quantity of 5 X plus 3 in quantity squared out of both of these terms. So doing so, this is going to be equal to the quantity of 5 X plus 3 in quantity squared, multiplied by the quantity, and in the first term, I'll have 2 multiplied by 5 X plus 3, and so that'll give me 10 X plus 6. Then for the second term, I'll have just plus 30 X. In quantity. And so now I just need to simplify that expression. So this is going to be equal to the quantity of 5 X plus 3 in quantity squared, multiplied by the quantity of 40 X plus 6 in quantity. So that is our first derivative. So now we need to find our second derivative. And so we're going to be looking for Y. And so this is going to be equal to the derivative with respect to x of our first derivative. So this could be the derivative of with respect to X of the quantity of 5 X plus 3 in quantity squared multiplied by the quantity of 40 X plus 6 in quantity. And so again, here, I have a product that I'm taking a derivative of. So we'll want to use our product rule again. So, in this case, we'll identify F of X. To be equal to the quantity of 5 X plus 3 in quantity squared, and Set G of X Equal to the second term, which is 40 X plus 6. So now we need to calculate F of X. And so this is going to be equal to the derivative with respect to X of quantity of 5 X plus 3 in quantity squared. And so again we'll have to apply our power rule, our chain rule, as well as our constant multiple rule and constant rule for derivatives. So this is going to be equal to. We'll first apply our power rule, so we'll have our exponent 2 multiplied by the quantity of 5 X plus 3 in quantity raised to the quantity of 2 minus 1, and this gets multiplied by the derivative with respect to x of the quantity of 5 X plus 3 in quantity by applying our chain rule. So this is going to be equal to. 2 multiplied by the quantity of 5 X plus 3 in quantity, and this gets multiplied by the derivative with respect to X of the quantity of 5 X plus 3, and we found earlier that the derivative of 5 X is just going to be 5, and the derivative of 3 is going to be 0, so we'll have 5 + 0 as our second term. And so simplifying this expression, this is going to be equal to 10 multiplied by the quantity of 5 X plus 3 in quantity. So now we need to find the derivative of G of X, so we'll look for G of X, and so it's gonna be equal to the derivative with respect to X. Of quantity of 40 x plus 6. In quantity. And so here we're just going to need to apply our constant multiple and power rule. So this is going to be equal to. 40, multiplied by 1, multiplied by X raised to the quantity of 1 minus 1 in quantity, plus 0. And so this simplifies 2, just being equal to 40. So now we can use our product rule to calculate our second derivative. So we'll have Y, and this can be equal to F of X, which is 10 multiplied by the quantity of 5 X plus 3 in quantity, and that gets multiplied by G of X, which is the quantity of 40 X plus 6 in quantity. Then we'll have plus. F of X, which is the quantity of 5 X plus 3 in quantity squared multiplied by G of X, which is 40. So now we just need to simplify this expression. So we can factor out a quantity of 5 x plus 3 in quantity out of both terms. So this is going to be equal to the quantity of 5 X plus 3 in quantity multiplied by the quantity, and here we'll multiply out the 10, multiplying the quantity of 40 X plus 6 in quantity for the first term. So we'll have 400 X. Plus 60. Then we'll have plus for the second term, we'll have the quantity of 5 X plus 3 in quantity multiplied by 40, which will give us 200 X. Plus 120. And so combined like terms, this is going to be equal to the quantity of 5 X plus 3 in quantity, multiplied by the quantity of 600 X plus 180 in quantity. And we can factor out a 60 out of the term of 600 X plus 180. And so, this is going to be equal to 60, multiplied by the quantity of 5 X plus 3 in quantity, multiplied by the quantity of 10 X plus 3 in quantity. And so this here is the answer that we're looking for. So I hope that this explanation has been useful, and I'll see you in the next video.

Second Derivatives

Find y'' in Exercises 59–64.

y = x(2x + 1)⁴

Key Concepts

Product Rule

Chain Rule