x^4 (x+h)^2*(x-h)^2/2
A arthur ohlsten Full Member Joined Feb 20, 2005 Messages 852 Mar 14, 2007 #2 y=x^4 y+h=[x+h]^4 expand y+h= = x^4+4x^3h+6x^2h^2+4xh^3+h^4 [y+h]-y= x^4+4x^3h+6x^2h^2+4xh^3+h^4-x^4 divide by h dy/dx=4x^3+ +6x^2h+4xh^2+h^3 let h-->0 dy/dx=4x^3 Arthur
y=x^4 y+h=[x+h]^4 expand y+h= = x^4+4x^3h+6x^2h^2+4xh^3+h^4 [y+h]-y= x^4+4x^3h+6x^2h^2+4xh^3+h^4-x^4 divide by h dy/dx=4x^3+ +6x^2h+4xh^2+h^3 let h-->0 dy/dx=4x^3 Arthur
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Mar 14, 2007 #3 Re: derivatives Hello, coop1024! I don't understand . . . x^4 (x+h)^2*(x-h)^2/2 Click to expand... How about some instructions ?? Are those two functions to be differentiated? . . (1) f(x) = x4\displaystyle (1)\;f(x)\:=\:x^4(1)f(x)=x4 . . (2) g(x) = (x+h)2(x−h)22\displaystyle (2)\;g(x)\;=\;\frac{(x+h)^2(x-h)^2}{2}(2)g(x)=2(x+h)2(x−h)2
Re: derivatives Hello, coop1024! I don't understand . . . x^4 (x+h)^2*(x-h)^2/2 Click to expand... How about some instructions ?? Are those two functions to be differentiated? . . (1) f(x) = x4\displaystyle (1)\;f(x)\:=\:x^4(1)f(x)=x4 . . (2) g(x) = (x+h)2(x−h)22\displaystyle (2)\;g(x)\;=\;\frac{(x+h)^2(x-h)^2}{2}(2)g(x)=2(x+h)2(x−h)2
