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y=[ln(1+e^x)]^6
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i know that u'/u but im not sure how to make it work with this problem
what is the question? take the derivative ?
yes
if the problem were y= u^6 you would do dy = 6u^5 du now replace u with ln(1+e^x) let 1+e^x be v you have u= ln(v) du = d (ln(v))= 1/v dv now if we replace u and v with the originals dy= 6 (ln(1+e^x))^5 * 1/(1+e^x) d (1+e^x) can you finish?
you use the chain rule, which might be confusing....
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yes that helped a bunch
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