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find dy/dx if y=(3x+2)^(ln(x))
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I've got ln(y) = ln(x)*ln(3x+2)
then i'm stuck
Then just use the product rule as normal.
you'll find that the derivative of lny= y'/y and since you just want y' in your final answer, just multiply both sides by y and plug in the y you originally started out with!
so, dy/dx (1/y) = (1/x)(ln 3x+2)+lnx(3/3x+2) dy/dx = (y)(1/x)(ln 3x+2)+lnx(3/3x+2)
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don't forget to multiply y times the second term as well and plug in your initial y (the function) from earlier so that everything is in terms of x.
dy/dx = (y) ((1/x)(ln 3x+2)+lnx(3/3x+2))
For your answer to be right, you must plug y back in, y=(3x+2)^(ln(x))
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