Asked by tom
find [dy/dx] x^2
Using this :
limit as the change of x —-> 0 for
f(x + deltax) - f(x)
—————————-
deltax
run through each step
Using this :
limit as the change of x —-> 0 for
f(x + deltax) - f(x)
—————————-
deltax
run through each step
Answers
Answered by
Damon
f(x+dx) = (x+dx)^2
= x^2 + 2 x dx + dx^2
f(x) = x^2
------------------------ subtract
f(x+dx) - f(x) = 2 x dx + dx^2
divide by dx
[ f(x+dx) - f(x) ] / dx = 2 x + dx
let dx ----> 0
= 2 x
the end
d/dx (x^2) = 2 x is now established in your mind forever.
= x^2 + 2 x dx + dx^2
f(x) = x^2
------------------------ subtract
f(x+dx) - f(x) = 2 x dx + dx^2
divide by dx
[ f(x+dx) - f(x) ] / dx = 2 x + dx
let dx ----> 0
= 2 x
the end
d/dx (x^2) = 2 x is now established in your mind forever.
Answered by
Damon
now if f(x) = x^n
f(x+dx) = (x+dx)^n = x^n + n x^(n-1) dx + [n(n-1)/2!]x^(n-2) dx*2 ..... + ....dx^bigger
f(x) = x^n still
-------------------------subtract
f(x+dx) -f(x) = n x^(n-1)dx+[n(n-1)/2!]x^(n-2) dx*2 .....
divide by dx
[f(x+dx) -f(x)] /dx = n x^(n-1) + [n(n-1)/2!]x^(n-2) dx ..... +...dx^ bigger
let dx ---> 0
d/dx(x^n) = n x^(n-1)
so for n = 5173
d/dx(x^5173) = 5173 x^5172 :)
f(x+dx) = (x+dx)^n = x^n + n x^(n-1) dx + [n(n-1)/2!]x^(n-2) dx*2 ..... + ....dx^bigger
f(x) = x^n still
-------------------------subtract
f(x+dx) -f(x) = n x^(n-1)dx+[n(n-1)/2!]x^(n-2) dx*2 .....
divide by dx
[f(x+dx) -f(x)] /dx = n x^(n-1) + [n(n-1)/2!]x^(n-2) dx ..... +...dx^ bigger
let dx ---> 0
d/dx(x^n) = n x^(n-1)
so for n = 5173
d/dx(x^5173) = 5173 x^5172 :)
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