Question

Can someone help me..
If f(x)=3x^2-x^3, find f'(1)

Answer 1

6x-3x^2

Answer 2

How'd you get that...

Answer 3

now plug in the one to get 6(1)-3(1)^2

Answer 4

oh are you just starting calculus I?

Answer 5

lol, yes...

Answer 6

then you probably havnt learned the shortcut

Answer 7

you used the power rule right?

Answer 8

nx^n-1?

Answer 9

yes

Answer 10

just bring the exponent down and multiply it by whats in front of x, then subtract 1 from numerater

Answer 11

But, my professor wants us to do it the longer way:/ or we won't get credit.. can you help me using that way?

f(a+h)-f(a)/h

Answer 12

yes i will one sec

Answer 13

so lets say you have f(x)=x^2

Answer 14

and you want to find f'(3)

Answer 15

so u start with \[\lim_{h \rightarrow infinity }\] of that equation you posted

Answer 16

\[f(x)=3x^2-x^3\]To find the derivative, we know that the derivative of a sum or difference is equivalent to the sum or difference of the derivatives of the terms.\[f'(x) = \frac{d(3x^2)}{dx}-\frac{d(x^3)}{dx}\\\ \ \ \ \ \ \ \ =6x-3x^2\]Now, evaluate the derivative for 1.\[f'(1)=6(1)-3(1)^2=6-3=3\]

Answer 17

does that help

Answer 18

if you have not got to the power rule yet, and you just want \(f'(1)\) you can use
\[\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}\] or
\[\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\] not sure what method you were taught

Answer 19

sometimes at the beginning it is easier to work directly with numbers

Answer 20

thanks guys :) you're so much help!

Answer 21

did i get a medal or whatever lol

Answer 22

\[f(x)=3x^2-x^3\]
\[f(1)=3-1=2\]
\[\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}\]
\[=\lim_{x\to 1}\frac{3x^2-x^3-2}{x-1}\]
\[=\lim_{x\to 1}\frac{-(x-1) (x^2-2 x-2)}{x-1}\]
\[=\lim_{x\to 1}-(x^2-2x-2)\]

Answer 23

power rule is undoubtedly easier but if you have to do it by hand you can write this

Answer 24

For the more rigorous proof...\[\lim_{\delta{x}\to\infty} \frac{f(x+\delta{x})-f(x)}{\delta{x}}\\=\lim_{\delta{x}\to\infty} \frac{3(x+\delta{x})^2-(x+\delta{x})^3-3x^2+x^3}{\delta{x}}\\=\lim_{\delta{x}\to\infty} \frac{3(x^2+2x\delta{x} + {\delta{x}}^2)-x^3-3x^2\delta{x}-3x{\delta{x}}^2-3x^2+x^3}{\delta{x}}\\=\lim_{\delta{x}\to\infty} \frac{3x^2+6x\delta{x}+3{\delta{x}}^2 - x^3 - 3x^2\delta{x} - 3x\delta{x}^2-3x^2+x^3}{\delta{x}}\\=\lim_{\delta{x}\to\infty} \frac{6x\delta{x} +3\delta{x}^2-3x^2\delta{x}-3x\delta{x}^2}{\delta{x}}\\=\lim_{\delta{x}\to\infty}[6x+3\delta{x}-3x^2-3x\delta{x}^]\\=6x-3x^2\]