jaychay Messages 58 Reaction score 0 Thread starter Oct 23, 2020 #1 Can you please help me ? I have tried to do it many times but I end up getting the wrong answer. Thank you in advance.
Can you please help me ? I have tried to do it many times but I end up getting the wrong answer. Thank you in advance.
skeeter Messages 1,103 Reaction score 1 Oct 23, 2020 #3 $\dfrac{d}{dx}[f(x)h(x)] = f’(x)h(x) + f(x)h’(x)$ $\displaystyle f(x)h(x) = \int f’(x)h(x) \, dx + \int f(x)h’(x) \, dx$ $\displaystyle (x+1)^5 = \sin{x} + \int f(x)h’(x) \, dx$ finish it ...
$\dfrac{d}{dx}[f(x)h(x)] = f’(x)h(x) + f(x)h’(x)$ $\displaystyle f(x)h(x) = \int f’(x)h(x) \, dx + \int f(x)h’(x) \, dx$ $\displaystyle (x+1)^5 = \sin{x} + \int f(x)h’(x) \, dx$ finish it ...
HOI Messages 921 Reaction score 2 Oct 21, 2021 #4 I think this is a problem on "integration by parts", NOT "partial differentiation".