Mark M. answered • 10/19/21
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Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Let y = (sin(2x))4x
Then, lny = ln[(sin(2x))4x] = 4xln(sin(2x))
So, y'/y = 4ln(sin(2x)) + 4x[2cos(2x)) / sin(2x)] = 4ln(sin(2x)) + 8xcot(2x)
Therefore, y' = (sin(2x))4x (4ln(sin(2x)) + 8xcot(2x))
Daniel K.
Thanks alot for the help on the problem and thanks doug for sending the link to the desmos it was very helpful
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10/19/21
Doug C.
Final answer needs an exponent of 4x on the (sin(2x)) Here is graph of f(x) and its derivative on Desmos, along with the general equation of tangent line at points on the graph of f. desmos.com/calculator/m9d4ifnh2f I was getting ready to answer the question, so just tagging along here.10/19/21