Question

When finding a derivative using the definition of a derivative, how do you know when the use the formula lim as y approaches x f(y)-f(x)/(y-x) vs. lim as h approaches 0 f(x+h)-f(x)/(h) ?

Answer 1

you can use either one

Answer 2

it is which ever one you like most

Answer 3

oh okay thanks

Answer 4

Both formulas are actually the same. Try replacing \(h\) with \(y-x\) to see what happens.

Answer 5

example:
\[f(x)=x^3 \\ f'(x)=\lim_{y \rightarrow x}\frac{f(y)-f(x)}{y-x}=\lim_{y \rightarrow x} \frac{y^3-x^3}{y-x}=\lim_{y \rightarrow x} \frac{(y-x)(y^2+xy+x^2)}{y-x} \\ f'(x)=\lim_{y \rightarrow x} (y^2+xy+x^2)=x^2+x(x)+x^2=3x^2 \\ \text{ or } f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \rightarrow 0}\frac{(x+h)^3-x^3}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}=\lim_{h \rightarrow 0} \frac{3x^2h+3xh^2+h^3}{h} \\ f'(x)=\lim_{h \rightarrow 0}(3x^2+3xh+h^2)=3x^2+3x(0)+0^2=3x^2\]

Answer 6

the first method involved factoring and the second one involved multiplying

Answer 7

it just depends on in that case if you preferred multiplication over the factoring
some people forget how to factor a difference of cubes

Answer 8

thought I hate multiplying sometimes

Answer 9

but I didn't really multiplied I sorta remember that line in the pascal's triangle :p

Answer 10

lol okay. thank you for the clear example. i just tried one myself using both methods and got the same answer.

Answer 11

sith mentioned proving they are equivalent

Answer 12

if you want to do that

Answer 13

or try that

Answer 14

\[(x-h)^3-x^3=([x+h]-[x])([x+h]^2+[x+h][x]+[x]^2) \\ (x+h)^3-x^3=h((x+h)^2+x(x+h)+x^2)\]
and another option for that second method
like we could have factored the numerator too