Find the derivatives of the functions in Exercises 19–40.

Question

y = (1 / 18)(3x − 2)⁶ + (4 − (1 / 2x²))⁻¹

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the derivative of Y is equal to 1 divided by 27, multiplied by the quantity of 9 X minus 5 in quantity cubed, plus the quantity of 6 minus 1 divided by the quantity of 3 x cubed in quantity in quantity raised to the -1 power. So here we need to determine the derivative of Y. So, we're looking for Y and that's gonna be equal to the derivative with respect to X. Of the quantity of 1 divided by 27, multiplied by the quantity of 9 X minus 5, in quantity cubed. Plus the quantity of 6 minus 1 divided by 3 multiplied by x raised to the minus 3 in quantity raised to the minus 1 in quantity. So here we've just rewritten 1 divided by X cubed as x raised to the minus 3 power. So in order to take this derivative, we're going to need to recall several derivative rules. And so the first one we're going to look at is going to be the constant multiple rule. So recall that if I have the derivative with respect to X of the quantity of C multiplied by X rates to the power n, that that is going to be equal to C multiplied by n, multiplied by X rays to the n minus 1. We also need to use our chain rule. So, call for the chain rule that we have the derivative with respect to X of F of G of X. And this is going to be equal to F of G of X. Multiplied by G of X. We'll also need to use our power rule, so we call for the power rule that we have the derivative with respect to X of quantity of X rated to the power N, and that is going to be equal to N multiplied by X rated to the quantity of N minus 1. And then finally we're going to need to use our constant rule, which is gonna be the derivative with respect to X of a constant C, and that is going to be equal to 0. So we're gonna need to use all these rules in order to take our derivative. So that means that our function Y is going to be equal to. For the first term, we'll use our um. Constant multiple rule. And so we'll have our constant, which is 1 divided by 27. And that's going to get multiplied by the exponent, which is 3. And then that's going to be multiplied by the quantity of 9 X minus 5 in quantity, and then that gets multiplied, or sorry, that gets raised to the power of 3 minus 1. But we also need to apply the chain rule to this function, so we'll need to multiply this by the derivative with respect to X of the quantity of 9 X minus 5 in quantity. Then we'll have plus for the second term, we're going to have to use our power rule. So we'll have the exponent, which is -1, multiplied by the quantity of 6 minus 1 divided by 3, multiplied by X raised to the -3 power in quantity, raised to the quantity of -1, 1 in quantity. And then we'll still have to apply the chain rule to this um term, so we need to multiply this by the derivative with respect to X. Of the quantity of 6 minus 1 divided by 3, multiplied by x raised to the minus 3 power in quantity. So simplifying this expression, this is going to be equal to. I'll have 3 divided by 27, which is 1 divided by 9. That gets multiplied by the quantity of 9 X minus 5 in quantity squared. Then I'll have to take the derivative with respect to x of the quantity of 9 X minus 5. And so, We're going for the first term, we're going to need to use our constant multiple rule again. So we'll have 9 multiplied by the exponent of our variable X, which is 1. That gets multiplied by X raised to the quantity of 1 minus 1 in quantity. Then we'll have minus the derivative of 5 with respect to X, and so there I'm taking the derivative of constant. So by our constant rule, that's going to be equal to 0 in quantity, then we'll have minus. The quantity of 6 minus 1 divided by 3, multiplied by x raised to the 3 power in quantity raised to the minus 2 power, and then that gets multiplied by the derivative with respect to X of the quantity of 6 minus 1 divided by 3, multiplied by x raised to the minus 3 power. And so again, for the first term we're going to be taking a derivative of a constant, so it's going to be equal to 0 by our constant rule. Then we'll have minus, and here for the second term, we'll have to use our constant multiple rule again. So we'll have our constant, which is 1 divided by 3, and that gets multiplied by our exponent, which is minus 3. And that gets multiplied by X rated to the quantity of -3, 1 in quantity. And so now we just need to simplify this a bit further. So this is going to be equal to. For the first term, I'll have 9 divided by 9, which is 1, so I'll just have the quantity of 9 X minus 5 in quantity squared. Then we'll have minus for the 2nd term. We're going to have the quantity of 6 minus 1 divided by 3, multiplied by X raised to the minus 3 power in quantity raised to the minus 2 power. That gets multiplied by, and here we'll have. -1 divided by 3, multiplied by -3, which is 1, and then that's multiplied by X raised to the -4. So this is gonna be multiplied by X rates to the -4 power. So now we can just rewrite everything to remove the negative exponents. And so doing so, this is going to be equal to. The quantity of 9 X minus 5 in quantity squared minus 1 divided by the quantity of X fourth multiplied by the quantity of 6 minus. 1 divided by the quantity of 3 x cubed. In quanti in quanti squared. And so this here is the answer that we are looking for for this problem. So I hope this explanation has been useful, and I'll see you in the next video.

Find the derivatives of the functions in Exercises 19–40.

y = (1 / 18)(3x − 2)⁶ + (4 − (1 / 2x²))⁻¹

Key Concepts

Power Rule

Chain Rule

Derivative of Reciprocal Function

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