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December 13th, 2018, 01:40 AM   #1
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Question 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?

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December 13th, 2018, 04:27 AM   #2
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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.
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December 13th, 2018, 07:30 AM   #3
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December 13th, 2018, 03:12 PM   #4
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Quote:
Originally Posted by gwilson View Post
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 $\displaystyle f'(x) = 1 + x^2$.

-Dan
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