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 December 13th, 2018, 12:40 AM #1 Newbie   Joined: Dec 2018 From: Sudan Posts: 1 Thanks: 0 Partial Differential Hello, if f is a function of x, how come this is true: I could not remember this rule of partial differentiation. Any help? Thanks December 13th, 2018, 03:27 AM #2 Newbie   Joined: Dec 2018 From: San Deigo Posts: 1 Thanks: 0 Math Focus: Algebra Notice that f(0+0)=f(0)+f(0) which implies that f(0)=0. Then, f′(0)=limh→0f(h)−f(0)h=limh→0f(h)h. Using f′(x)=limh→0f(x+h)−f(x)h=limh→0f(x)+f(h)+x h(x+h)−f(x)h So f′(x)=limh→0f(h)h+limh→0xh(x+h)h=f′(0)+x2 So f′(x)=−1+x2 Finally, f′′(x)=2x. December 13th, 2018, 06:30 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,751 Thanks: 2135 The image should be uploaded instead of linked to. Thanks from topsquark December 13th, 2018, 02:12 PM   #4
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 Originally Posted by gwilson Notice that f(0+0)=f(0)+f(0) which implies that f(0)=0. Then, f′(0)=limh→0f(h)−f(0)h=limh→0f(h)h. Using f′(x)=limh→0f(x+h)−f(x)h=limh→0f(x)+f(h)+x h(x+h)−f(x)h So f′(x)=limh→0f(h)h+limh→0xh(x+h)h=f′(0)+x2 So f′(x)=−1+x2 Finally, f′′(x)=2x.
Are you trying to help solve a different question? We aren't given an f(x) so we can't say f(0 + 0) = f(0) + f(0) nor can we say that $\displaystyle f'(x) = 1 + x^2$.

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