In Exercises 7–38, find the derivative of y with respect to x,...

Question

In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.

18. y = t√(ln t)

Accepted Answer

Welcome back everyone. Find the derivative of Y given that Y equals X multiplied by square root of LN X2. For this problem, we are going to identify Y, which is a derivative of Y, and we have to apply the product rule because we have a product of two functions. According to the product rule, we take the derivative of X, we multiply by square root of LN X2, and we're adding X, which is multiplied by the derivative of. Square root of LN of X2, right? So essentially what we want to do from here is identify each derivative. Or simplicity we can express square root of LM of X2d as LN of X2. Erase the power of 1/2 so that we can apply the power rule, right? Now the derivative of X is equal to 1. And how to differentiate a land of X2? Raising the power of one half, we are going to apply the. General Before we begin, we are going to use the properties of logarithms. We can express natural logs to Alanoff X. And now applying the chain rule, we're going to get 1/2 multiplied by 2 LN of X raised to the power of -12 multiplied by the derivative of 2 LN of X based on the chain rule we're multiplying by the derivative of the inner function, right? So now let's simplify. This is 1/2 multiplied by 2 LN of X raises the power of negative 1/2 multiplied by. 2 divided by X because the derivative of LN of X is 1 divided by X. And now we can cross out. The 2 And we can express. This derivative is one divided by X. Multiplied by square root of two L and of X. Now, what we're going to do is also take into account the domain, right? X is not equal to 0. So we're going to write it in the correct form. That would be 1 divided by X. Square root of L N X squared. And now let's go ahead and substitute this derivative into the product rule. Y is going to be equal to 1 multiplied by square root of LN X2 + x multiplied by 1 divided by X square root of LN X2. We can cancel out x divided by X, and we can use the common denominator now. So the common denominator is square root of LN X2. If we multiply it by square root of LN X2, we're going to get LN X2, and we are adding 1. This is how we end up with the final derivative, which is LN X2 + 1 divided by square root of LN X2. Thank you for watching.

In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.18. y = t√(ln t)

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In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.
18. y = t√(ln t)