75–86. Logarithmic differentiation Use logarithmic differentia...

Question

75–86. Logarithmic differentiation Use logarithmic differentiation to evaluate f'(x).

f(x) = (x+1)^3/2(x-4)^5/2 / (5x+3)^2/3

Accepted Answer

Hi everyone, let's take a look at this practice problem dealing with derivatives. This problem says to use logarithmic differentiation to find the derivative of F of X, which is equal to the quantity of 2 X plus 2 in quantity rates to the 4th power, multiplied by the quantity of 3x minus 6 in quantity cubed, divided by the quantity of 4 X plus 1 in quantity rates to the 5th power. So we need to find our derivative of F of X using logarithmic differentiation. The first thing we want to do is rewrite our function. So we're going to have F of X. is equal to and in the first term of the numerator, I'm going to factor out a 2 rates to the 4th power. So I have 2 rates to the 4th power, multiplying what's left, which is going to be the quantity of X + 1 in quantity rates to the 4th power. I have the second term, I'm going to factor out a 3 cubed, so I'll multiply this by 3 cubed. And that gets multiplied by what is left, and that is going to be the quantity of X minus 2 in quantity cubed, and that gets divided by the quantity of 4 X plus 1 in quantity rates to the 5th power. So simplifying the numerator, this is going to be equal to 432, multiplied by the quantity of X + 1 in quantity rates to the fourth power, multiplied by the quantity of X minus 2 in quantity cubed, divided by the quantity of 4 X + 1. In quanti raised to the 5th hour. So, we need to perform logarithmic differentiation. So the first thing I need to do is take the natural log of our function F of X. So we'll have the natural log of F of X. is equal to the natural log of the quantity of 432 multiplied by the quantity of X + 1 in quantity rates to the fourth power multiplied by the quantity of X minus 2 in quantity cubed divided by the quantity of 4 x + 1 in quantity raised to the 5th power. So we're going to rewrite the right hand side of this equation using our properties of natural logs. And we're mainly gonna be using. Our power, product, and quotient rule for logarithms. So we'll have the natural log of FF X. is equal to the natural log of 432. Plus, or multiplied by the natural log of the quantity of X plus 1 in quantity. Plus, 3 multiplied by the natural log of the quantity of X minus 2 in quantity. Minus 5 multiplied by the natural log of the quantity of 4 X plus 1 in quantity. So now we need to do is take the derivative with respect to X of both sides of this equation. So doing so on the left hand side of this equation, recall that when I take the derivative of the natural log, I get one divided by its argument, which is to be 1 divided by F of X. But we'll need to apply a chain rule here, so we'll need to multiply this by the derivative of the argument, which is going to be F of X. And so this is going to be equal to. Well, on the right hand side we'll have the derivative of the natural log of 432 with respect to X. And here I'm taking the derivative of a constant. So, when I take a derivative of a constant, it's derivative is always going to be zero, so I have 0, plus. Or, multiplied by the derivative with respect to X of the natural log of the quantity of X + 1. And so when I take that derivative, I'm gonna get 1 divided by the quantity of X + 1 in quantity, multiplied by the derivative of X + 1, which is just 1. Then we'll have +3, multiplied by the derivative of the natural log of the quantity of X minus 2. And so we want to take that derivative and we'll get 3 multiplied by the quantity of 1 divided by X minus 2. In quantity, multiplied by the derivative of X minus 2, which is just 1. Then I'll have minus 5 multiplied by the derivative of the natural log of the quantity of 4 X + 1. So taking that derivative, I'm going to get 1 divided by 4 X plus 1. In quantity, multiplied by the derivative of 4 X plus 1, which is just 4. So simplifying the right hand side of that equation, we're going to get 1 divided by F of X. Multiplied by F of X. is equal to 4 divided by the quantity of X plus 1 in quantity plus 3 divided by the quantity of X minus 2 in quantity minus 20 divided by the quantity of 4 X plus 1 in quantity. So we'll want to combine all of our terms on the right hand side. So doing so, I will have one divided by F of X. Multiplied by F of X. is equal to 4 multiplied by the quantity of X minus 2 in quantity, multiplied by the quantity of 4 X plus 1 in quantity. Plus 3 multiplied by the quantity of X plus 1 in quantity, multiplied by the quantity of 4 X plus 1 in quantity. Minus 20 multiplied by the quantity of X + 1 in quantity, multiplied by the quantity of X minus 2 in quantity, and all of that is divided by The product of the quantity of X + 1 in quantity, multiplied by the quantity of X minus 2 in quantity, multiplied by the quantity of 4 X plus 1 in quantity. So multiplying everything out in our numerator, we're going to get one divided by F of X. Multiplied by F of X. is equal to 16 X squared. Minus 28 X. -8. Plus 12 X squad. Plus 15 X. Plus 3. Minus 20 x 2. Plus 2 X. Plus 40, and all of that is divided by the product of the quantity of X plus 1 in quantity, multiplied by the quantity of X minus 2 in quantity, multiplied by the quantity of 4 X + 1 in quantity. So, now what we want to do is solve this equation for F of X. So, I'll do that by multiplying both sides of our equation by F of X. So I'll have F of X. is equal to F of X. Multiplied by the quantity, and here we'll simplify our numerator by collecting like terms. So we'll have 16 X2 plus 12 X2 minus 20 X2, that will give me 8 X2. Then if I look at my linear terms, I'll have -28 X. Plus 15 X plus 20 X. So then we'll have plus 7 X as our next term in our numerator. Then if I look at The constant terms will have -8. Plus 3. Plus 40, and so that gives me plus 35 as my constant term in my numerator, and all of that is divided by the product of the quantity of X + 1 in quantity, multiplied by the quantity of X minus 2 in quantity, multiplied by the quantity of 4 X plus 1 in quantity. So now what's the substitute in our expression for F of X, so we'll have F of X. is equal to 432 multiplied by the quantity of X + 1 in quantity raised to the fourth power, multiplied by the quantity of X minus 2 in quantity, divided by. The quantity of 4 X plus 1 in quantity raised to the 5th power, and that fraction is multiplied by the quantity of 8 X2 + 7 X. Plus 35 in quantity, divided by the product, which is the quantity of X + 1 in quantity, multiplied by the quantity of X minus 2 in quantity, multiplied by the quantity of 4 X plus 1 in quantity. So now all we need to do is to simplify our expression. So doing so, we will have F of X. is equal to 432. Multiplied by the quantity of X + 1 in quantity cubed. Multiplied by the quantity of X minus 2. Squared multiplied by the quantity of 8 X2. Plus 7? X + 35. In quantity? Divided by The quantity of 4 X plus 1 in quantity raised to the 6th power. And so, this is the derivative of our function F of X that we were looking for. So this is the answer to our problem. So I hope that this explanation has been useful, and I'll see you in the next video.

75–86. Logarithmic differentiation Use logarithmic differentiation to evaluate f'(x).
f(x) = (x+1)^3/2(x-4)^5/2 / (5x+3)^2/3

Key Concepts

Logarithmic Differentiation

Product and Quotient Rules

Chain Rule

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