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Find the derivative of the function.

$ y = e^{\sin 2x} + \sin (e^{2x}) $

$2 e^{\sin 2 x} \cdot \cos 2 x+2 e^{2 x} \cos \left(e^{2 x}\right)$

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Missouri State University

Oregon State University

Baylor University

Boston College

here we have an interesting function. I suspect the problem writer had fun putting this one together. We have different functions. Will same functions in both Halfs of problem just arranged in different ways. So let's use the chain rule on each term. So to find the derivative of each of the sign of two X, we start with you know how the derivative of either the exes e to the X. So the derivative of either the sign to act starts with each of the sign to X, and then we multiply by the derivative of sign two X and signed two X has an outer layer sign and an inner layer two X So the derivative of the sign part gives us co signs with co sign of two X and the derivative of the two ex parte gives us too. So that was a three layer chain rule problem. And now we move on to the second term. We have the sine function and inside that we have the each of the X function and inside that we have two x so another three layer problem. So we start with the derivative of the outside sign and So it's derivative is co sign of co sign of each of two X, and then we multiply by the derivative of the each of the two X, which would be each of the two x Times two. Now we're looking for ways to simplify this. There's really nothing that combines weaken. Just rewrite the parts in a different order, so we can write to each of the sign of two x Times Co sign of two x. Sorry about that technical difficulties, plus to E to the two x Times Co sign of each of the two x.

Oregon State University