In Exercises 57–70, use logarithmic differentiation to find th...

Question

In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.

70. y = ∛(x(x+1)(x-2)/(x²+1)(2x+3))

Accepted Answer

Welcome back everyone. In this problem we want to use logarithmic differentiation to find the derivative of y with respect to x, for y equals the 4th root of x multiplied by x + 2 multiplied by 2 x minus 3, divided by x2 + 4 multiplied by 3 x + 5. Now for us to use logarithmic differentiation, we'll somehow have to introduce logs into our equation, right, and we can do that by taking the natural logs off both sides. It will be easier if we rewrite our expression or rewrite our equation. Y equals the 4th root of all that. Expression that we have underneath our 4th root, it's quite the mouthful. So we'll have this. And we'll rewrite it as. X multiplied by x + 2 multiplied by 2 x minus 3. Multiplied by x2 + 4 to the -1. Multiplied by 3 x + 5 to the -1, all raised to the power of 1/4 and now it will be easier for us to apply our logs. Our log or logs will be more useful. How so? Well, recall one property of logarithms is that if we're finding the log of a product, say, the natural log of A B C, we can rewrite it as LNA plus LNB plus LNC. On the other hand, we also know that if we're finding the log of a term raised to a power like the natural log of A to the power of B, we can bring down our power and rewrite it as B L N A. So with these two properties we can use it to simplify our entire expression, meaning we can bring down our powers wherever we can in the logarithmic term, and we can rewrite our products as sums for our logs. So in this case when we take the natural log off the left hand side we'll get LNY and for the right hand side we bring down our power to get a 4th and then we can rewrite it as LNX plus LNX + 2. Plus LN 2 x minus 3. Minus LNX2 + 4, because in this case we're we're raising it to the power of -1. Minus 1 of 3 x + 5 Now that makes it easier for us to apply logarithmic differentiation because we can differentiate implicitly. Yeah, differentiating implicitly. With respect to x. Now on our right hand side we'll end up with 1 divided by y multiplied by D Y D X, but no, on our left, sorry, that's on our left hand side, but on our right hand side we have quite a bit of differentiation to do. We need to differentiate each term. No, the derivative of LNX will be 1 divided by X. For LNX + 2, it's going to be 1 divided by X + 2. For the LN of 2 X minus 3, it's going to be 2 divided by 2 x minus 3. For a negative LN of X2 + 4, we're going to end up with -2 x divided by X2 + 4. And for a negative LN of 3 x + 5, we'll end up with -3 divided by 3 x + 5. So these are the derivatives of all of our terms. Now we have DYD X. We just need to solve for it. Now we can multiply both sides by y, and when we do that, we get DYD X. To be equal to 1/4 of Y, OK. Multiplied by this expression and let me just uh bring it down to our next line in the interest of time here. But now what do we know about y? Remember, originally we were given the value of y or what y is equal to in our problem statement. So if we rewrite that here, then we end up with the derivative being 1/4 of the 4th root of x multiplied by x + 2 multiplied by 2 x minus 3 divided by x2 + 4 multiplied by 3 x + 5. OK. All multiplied by 1 divided by x plus 1 divided by x + 2 + 2 divided by 2 x minus 3 minus 2 x divided by x2 + 4 minus 3 divided by 3 x + 5. This is the derivative of Y with respect to X using logarithmic differentiation. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.70. y = ∛(x(x+1)(x-2)/(x²+1)(2x+3))

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In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.
70. y = ∛(x(x+1)(x-2)/(x²+1)(2x+3))