f(x)= (e^x +1)^2 I am not sure how to find the derivative of this one! Do I (e^x +1)(e^x +1)? Thanks in advance!
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Use the chain rule. du^2/dx = 2u du/dx Where u = e^x + 1
Originally Posted by advancedfunctions2010 f(x)= (e^x +1)^2 I am not sure how to find the derivative of this one! Do I (e^x +1)(e^x +1)? Thanks in advance! $\displaystyle f(x) = (e^x + 1)(e^x + 1)$ $\displaystyle f'(x) = (e^x)(e^x + 1) + (e^x + 1)(e^x) = 2e^{2x} + 2e^x$ Or I suppose you could have said: $\displaystyle f(x) = e^{2x} + 2e^x + 1$ $\displaystyle f'(x) = 2e^{2x} + 2e^x + 0 = 2e^{2x} + 2e^x$
Or simply use the chain rule as snowtea suggested: $\displaystyle f'(x)= 2(e^x+ 1)(e^x+1)'= 2(e^x+ 1)(e^x)= 2e^{2x}+ 2e^x$
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