Question

I have a question !!

Answer 1

So we have a constant.

\(\large\color{#000000}{\displaystyle f(x)=k}\)

\(\large\color{#000000}{f'(x)=\displaystyle\lim_{h \rightarrow ~0}\frac{f(x+h)-f(x)}{h}}\)

\(\large\color{#000000}{\displaystyle f'(x)=\lim_{h \rightarrow ~0}\frac{k+h-k}{h}=\lim_{h \rightarrow ~0}\frac{h}{h}=1}\)

Answer 2

How come is the answer 1 ??

Answer 3

\[\Large \lim_{h \to 0}\left(\frac{h}{h}\right) = \lim_{h \to 0}\left(1\right) = 1\]

I'm using the rule

\[\Large \lim_{x \to a}\left(k\right) = k\]
a,k are constants

Answer 4

Yes, but this limit is the derivative of constant k with respect to x....

Answer 5

I am guessing that this definition of differentiation is not going to work for a constant function (((afterall, a slope of y(x)=C is 0 at x=a for all a)))

Answer 6

why wouldn't it work? the constant function is continuous everywhere, so it's continuous at x = a

Answer 7

Yes, but if we get the result that the derivative of a constant is 1, then there is something wrong

Answer 8

But, using this limit definition, without all properties of limits, we get
d/dx (k)=1

Answer 9

oh right I'm not thinking

Answer 10

somehow missed that you should have k-k in the numerator

NOT k+h-k

Answer 11

oh right

Answer 12

f(x) = k
f(x+h) = k

Answer 13

yes, yes

Answer 14

I was plugging into k as if it x.... tnx for pointing it out

Answer 15

So we have a constant.

\(\large\color{#000000}{\displaystyle f(x)=k}\)

\(\large\color{#000000}{f'(x)=\displaystyle\lim_{h \rightarrow ~0}\frac{f(x+h)-f(x)}{h}}\)

\(\large\color{#000000}{\displaystyle f'(x)=\lim_{h \rightarrow ~0}\frac{k-k}{h}=\lim_{h \rightarrow ~0}\frac{0}{h}=0}\)

Answer 16

:)))
tnx

Answer 17

np