find the derivative of f(x) = 7x^3. I think the answer should be 21x^2 but I'm getting 7x^2. I will put my workings underneath.

(cf)'=cf' you know this rule? f'(x)=7(x^3)'=7*3x^2=21x^2

7x^3 - 7(x + h)^3/h - this may be the problem 7x^3 - 7x^3 + 7x^2h + 7xh^2 + 7h^3/h - or this 7x^2 + 7xh + 7h^2 make h 0 7x^2

also do you know (x^n)'=nx^(n-1) power rule

Yeah I am thinking that I am wrong because when I apply the shortcut version it is wrong. But I want to understand how to do it in the long sense so I can do hard ones later..

Oh I only know your second rule, not the first one

k f(x)=7x^3 f(x+h)=7(x+h)^3=7(x^3+3x^2h+3xh^2+h^3)=7x^3+21x^2h+21xh^2+7h^3

\[\frac{(7x^3+21x^2h+21xh^2+7h^3)-(7x^3)}{h}\]

\[\frac{21x^2h+21xh^2+7h^2}{h}\]

Ahhh... again the binomial theory defeats me! I didn't put the 3's when I expanded it... Thankyou!

\[\frac{h(21x^2+21xh+7h)}{h}\]

k so you got the rest?

I think I'm good, I see where I went wrong now :) Thanks :)

:)

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