In Exercises 1–24, find the derivative of y with respect to th...

Question

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.

19. y = t arctan(t) - 1/2 ln(t)

Accepted Answer

Welcome back, everyone. Find D Y divided by D T for Y equals T multiplied by inverse of tangent of C minus 1/3 LN of C. For this problem we want to find the derivative of Y. We can just use Y as a shorthand notation. Let's recognize that we have a difference of two functions, and the first function is actually a product of two functions. So what we're going to do for the first part is apply the product rule of differentiation. Let's remember that first of all, we take the derivative of the first function, so the derivative of T. We're going to multiply by the second function which is inverse of tangent of T, and we are going to add the first function C multiplied by the derivative of the second one, so the derivative of inverse of tangent of C. So this is how we apply the product rule for the first part. Now we're going to subtract 1/3 multiplied by the integral of LN of T, right? Because we can factor out the constant. And now, as we can see, we essentially want to evaluate three derivatives. The derivative of T is 1, right? DT divided by DT. We also want to get the derivative of Inverse of tangent. It is an elementary derivative, and its value is equal to 1 divided by 1 plus T2. And finally, we have another elementary derivative, that's the derivative of A lot of tea, right? Its value is equal to 1 divided by T. And now what we can do is simply substitute all of these into our formula, so we get 1 multiplied by inverse of tangent. So we're going to write inverse subtangent of T. Plust multiplied by the derivative of inverts of tangent that's going to be t divided by. 1 + T squared and we're going to subtract 1/3 multiplied by 1 divided by T. So that's a 1 divided by 3 T. and that's our final answer for this problem. Thank you for watching.

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.19. y = t arctan(t) - 1/2 ln(t)

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In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
19. y = t arctan(t) - 1/2 ln(t)