In Exercises 13–24, find the derivative of y with respect to t...

Question

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.

15. y = 2√t tanh(√t)

Accepted Answer

Welcome back, everyone. Find the D Y divided by D T 4 Y equals 4 square root of T multiplied by cash of square root of T. For this problem we want to find the derivative of Y with respect to C. We have to notice that we have a product of two functions. Additionally, we can factor out the constant, which is for using the properties of derivatives, and we can now apply the product rule for the remaining product. So, according to the product rule, we want to take the derivative of the first function, so that's the derivative of square root of T. It is multiplied by the second function, which is cash of square root of T, and we're going to add the first function, square root of T, multiplied by the derivative of the second one. Gosh of squirt of tea. According to this expression, we want to get two derivatives. One of them is square root of T. We can rewrite it as T raises the power of 1/2 and apply the power rule. The derivative is 1/2 T to the power of -12. We can now rewrite it as 1 divided by 2 to the power of positive 1/2, which is basically 1 divided by 2 square root of T. And this is how we get our first derivative, right. Now, for the other derivative, we have to understand that it is a composite function. Cash of square root of T derivative is going to be found using the chain rule. We want to differentiateca with respect to the inner function, and the derivative of cash is sing. So first of all, we get singe of square root of T. And based on the chain rule, we're multiplying by the derivative of the inner function. So the derivative of square root of T, which we already have. So the overall product here is going to be singe. Of square root of tea. Divided by 2 square roots of T. And now that we have both derivatives, we can substitute into the product rule to get. Or multiplied by. In rents we have cash. Of square root of tea. Which is divided by the derivative of square root of T, that's 2 square root of T, right? And we are adding square root of T. Multiplied by the derivative of cache of squiro of T which is. Sen Square root of tea. Divided by 2 square root of tea. We can simplify slightly. We can cross out square root of t for one of our terms, and we can distribute as well. Or divided by 2 gives us 2, so we get 2. Multiplied by. Gosh Of square root of T. Divided by Square root of tea. And then we're adding. Singe multiplied by 4 divided by 2. This gives us 2, right? So we can just add 2. Sin Square root of tea. Additionally, we can rationalize the denominator of the first part. So instead of having square root of T, we can write T and add square root of T in the numerator. So, the final answer is going to be. 2 square root of tea, cache of square root of tea. Divided by T plus 2 cent square root of T. Thank you for watching.

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.15. y = 2√t tanh(√t)

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In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
15. y = 2√t tanh(√t)