05 Nov '04 21:59

i know that the derivative of f(x)=x^2 is f'(x)=2x

but if we put that in the form of mx+b, where m=the slope, then the coefficent on x^1 is x, so m=x ---->the dreivative.

clearly this is not the case, by why? it makes a lot of sence

another example of my point:

f(x)=x^4-2x^3+3x^2-4x+5

f'(x)=m=4x^3-6x^2+6x-4

but in the form mx+b, f(x)=(x^3-2x^2+3x-4)x+5

so why dosnt m=f'(x)=x^3-2x^2+3x-4 ?

but if we put that in the form of mx+b, where m=the slope, then the coefficent on x^1 is x, so m=x ---->the dreivative.

clearly this is not the case, by why? it makes a lot of sence

another example of my point:

f(x)=x^4-2x^3+3x^2-4x+5

f'(x)=m=4x^3-6x^2+6x-4

but in the form mx+b, f(x)=(x^3-2x^2+3x-4)x+5

so why dosnt m=f'(x)=x^3-2x^2+3x-4 ?