f(x+h) = 5(x+h)^2 - 3(x+h) + 1
= 5x^2+10xh+5h^2 -3x-3h + 1
f(x) = 5x^2 -3x+1
subtract
f(x+h) - f(x) = 10 xh + 5 h^2 -3 h
divide by h to get slope
slope = 10 x +5 h - 3
let h go to zero to get slope at x
dy/dx = 10 x -3
now try the other one
I understand finding the derivative and there is a simpler way of doing so, however, can someone show you derive at these answers by the DEF'N of Derivatives? Much appreciated!
Question 1: Find, from the definition, the derivative, f'(x), for f(x)=5x^2-3x+1 .
Answer: f ’(x) = 10x-3 .
Question 3: Find, from the definition, the derivative, f'(x), for f(x)=√(9x+4) .
Answer: f ’(x) = 9/(2√(9x+4)) .
3 answers
f(x+h) = (9x+9h+4)^.5
[f(x+h)]^2 = 9 x + 9 h + 4
[f(x)]^2 = 9x+4
subtract and get 9h, remember that
a^2-b^2 = (a+b)*(a-b)
so
[f(x+h)]^2 - [f(x)]^2 =
{[f(x+h)]+[f(x)]}*{[f(x+h)]-[f(x)]}
SO
{[f(x+h)]-[f(x)]}={[f(x+h)]^2- [f(x)]^2} /{[f(x+h)]+[f(x)]}
ah ha, I want what I see on the left
now from way up there I know
{[f(x+h)]^2- [f(x)]^2} = 9h
so
{[f(x+h)]-[f(x)]}=9h /{[f(x+h)]+[f(x)]}
divide by h to get slope
9 /{[f(x+h)]+[f(x)]}
let h --->0
9/{2 f(x)}
= 9/{2sqrt(9x+4)}
so I agree with you
[f(x+h)]^2 = 9 x + 9 h + 4
[f(x)]^2 = 9x+4
subtract and get 9h, remember that
a^2-b^2 = (a+b)*(a-b)
so
[f(x+h)]^2 - [f(x)]^2 =
{[f(x+h)]+[f(x)]}*{[f(x+h)]-[f(x)]}
SO
{[f(x+h)]-[f(x)]}={[f(x+h)]^2- [f(x)]^2} /{[f(x+h)]+[f(x)]}
ah ha, I want what I see on the left
now from way up there I know
{[f(x+h)]^2- [f(x)]^2} = 9h
so
{[f(x+h)]-[f(x)]}=9h /{[f(x+h)]+[f(x)]}
divide by h to get slope
9 /{[f(x+h)]+[f(x)]}
let h --->0
9/{2 f(x)}
= 9/{2sqrt(9x+4)}
so I agree with you
I did the second one because it is much harder than the first one and if you had trouble with the first you were in much greater trouble with the second.
2 answers