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

Question

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

f(x) = x⁸cos³ x / √x-1

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 the function F of X, which is equal to x 5 multiplied by sin squared of x, divided by the square root of the quantity of X plus 7. Now we're asked to find the derivative of the function F of X using logarithmic differentiation. So first step is to take the natural log of our function. So we'll have the natural log of F of X. is equal to the natural log of the quantity of X 5 multiplied by sin squared of x divided by the quantity of X + 7 in quantity raised to the 1/2 hour. So here we've just rewritten the square root of quantity of X + 7 in quantity as the quantity of X + 7 in quantity rates to the 1/2 power. So now we're going to rewrite the right hand side of this equation using our properties of logarithms. Mainly we're going to use our power rule, product rule, and quotient rule for logarithms. So doing so, we'll have the natural log of F of X. is equal to 5 multiplied by the natural log of X. Plus 2 multiplied by the natural log of sin of X. Minus 1/2 multiplied by the natural log of the quantity of X + 7 in quantity. So now what 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, recall that when you take the derivative of the natural log function, you get 1 divided by the argument, which in this case will be 1 divided by F of X, but we'll need to apply our train rule, so we'll need to multiply this by the derivative of the argument. So when I take the derivative of the argument, I'll have the derivative of F with respect to X, which is going to be F of X. And then this is going to be equal to 5 multiplied by the derivative with respect to X of the natural log of X. So, taking the derivative of the natural log of X, I get 1 divided by X. Then we'll have plus 2 multiplied by the derivative with respect to X of the natural log of sin of X. So again, taking the derivative of the natural log function, I'll get 1 divided by the argument. We'll have 1 divided by sine of X. But I'll need to apply a chain rule, so I'll need to multiply this by the derivative of sine of X with respect to X, which is cosine of X. Then I'll have minus 1/2 multiplied by the derivative with respect to X of the natural log of the quantity of X + 7. So again, taking the derivative of the natural log function, I'll get 1 divided by the argument, which is 1 divided by X + 7, and we'll have to multiply this by the derivative of the argument, again applying our chain rule. So the derivative with respect to X of the quantity of X plus 7 is just 1, so we'll multiply this by 1. So now we want to just simplify on the right-hand side of our equation. So we'll have one divided by F of X. Multiplied by F of X. is equal to 5 divided by X. Plus 2 multiplied by cotangent of X. So recall that cosine of X divided by sin of X is equal to cotangent of X. Then we'll have -1 divided by the quantity of 2 multiplied by the quantity of X plus 7 in quantity. So, we want to combine all of our terms on the right hand side under a single, uh, under a common denominator. So we'll have 1 divided by F of X. Multiplied by F of X. It's equal to And here in the numerator we'll have 5 multiplied by 2 multiplied by the quantity of X + 7 in quantity, plus 2 multiplied by X, multiplied by 2 multiplied by the quantity of X plus 7 in quantity multiplied by cotangent of X. Minus X, and all of this is divided by the quantity of 2 X multiplied by the quantity of X plus 7. So multiplying everything out in the numerator on the right hand side, we will get 1 divided by F of X. Multiplied by F of X. is equal to 10 X Plus 70 Plus 4 X 2 multiplied by cotangent of X. Plus 28 X multiplied by cotangent of X. Minus X. And that whole quantity is divided by the quantity of 2 X multiplied by the quantity of X plus 7 in quantity. So now what we need to do is solve this equation for F of X. So to do that, we'll multiply both sides of the equation by F of X. So we'll have F of X. is equal to F of X. Multiplied by the quantity, and here we'll just collect like terms in our new mer. We'll start off with our quadratic terms. We only have one term, which is 4 X squad, multiplied by cotangent of X. Then we'll have plus for our next term. We'll look at terms that just have X and cotangent of X in them. We only have one, we'll have plus 28 X multiplied by cotangent of X. Next, we'll look at terms that just have X in them, so we have 10 X minus X, so we'll have plus 9 X. Then we'll look at a constant terms, and we just have 1, we'll have plus 70, and that whole quantity is divided by the quantity of 2 X multiplied by the quantity of X plus 7 in quantity. So now we need to substitute in our um function that we originally had for FF X. So we'll have F of X. is equal to for F of X, recall that that was x 5 multiplied by sine squared of x divided by the square root of the quantity of X plus 7 in quantity, and that gets multiplied by the quantity of 4 X 2 multiplied by cotangent of X. Plus 28 X multiplied by cotangent of X plus 9 X plus 70 in quantity, divided by the quantity of 2 X multiplied by the quantity of X plus 7 in quantity. So now we just need to simplify this expression. So that means I'll have F of X is equal to. I can cancel out one factor of X with the X in the denominator. So we'll have X to the 4th, multiplied by sine squared of X, multiplying the quantity of 4 x 2 multiplied by cotangent of X plus 28 X multiplied by cotangent of X. Plus 9 X. Plus 70 in quantity, divided by. The quantity of 2 multiplied by the quantity of X plus 7 in quantity raised to the 3 halves power. So this here is our final answer for the problem. So here we have calculated our derivative of our function F of X, which is F of X. 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) = x⁸cos³ x / √x-1

Key Concepts

Logarithmic Differentiation

Product and Quotient Rules

Chain Rule

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