Find n^( th ) derivative of 2^(x)sin^(2)x cos x
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Answer:
Given : y=e2x×cos2x×sinx and I have to find it's nth derivative
I have managed to break y down to:
y=14×(e2x.sin3x+e2x.sinx)
But I don't know how to apply Leibnitz Rule in cases like (eax×sinbx).
Step-by-step explanation:
Hope it helps you
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