Question

How do I find the derivative of this function using the definition of a derivative
? f(x)=4x^2-3x^2+2x-pi
Is finding the derivative of a function basically plugging the function into the formula then simplifying it?

Answer 1

Derivative of a function can be found by using the below formula
\[\frac{ d }{ dx}[x^n] = nx^{n-1}\]

Example \[\frac{ d }{ dx }[4x^2] = 2*4x^{2-1} = 8x\]

Answer 2

on our test we cannot use differential formulas we have to solve the derivative using the definition which is f(x)-f(a)/x-a. The only other tool we can use is limit laws

Answer 3

Okay then you got to use the below definition of derivate

\[\frac{dy}{dx}[x]= \lim_{h \rightarrow \infty}\frac{f(x+h)-f(x)}{h}\]

Answer 4

well yes, I plug it in and I get this far :

\[\frac{( 4x^3-3x^2+2x-\pi +h)-4x^3+3x^2+2x-\pi}{ h }\]

and I made a typo, the first part is 4x^3 not 4x^2

Answer 5

I don't know how to simplify is further or how to find the derivative of each individual part

Answer 6

Woops you didn't plug it in correctly,\[\large\rm f(\color{orangered}{x})=4(\color{orangered}{x})^3-3(\color{orangered}{x})^2+2(\color{orangered}{x})-\pi\]So when evaluated at x+h,\[\large\rm f(\color{orangered}{x+h})=4(\color{orangered}{x+h})^3-3(\color{orangered}{x+h})^2+2(\color{orangered}{x+h})-\pi\]

Answer 7

So then,\[\large\rm f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]gives us,\[\rm f'(x)=\lim_{h\to0}\frac{[4(x+h)^3-3(x+h)^2+2(x+h)-\pi]-\left[4x^3-3x^3+2x-\pi\right]}{h}\]

Answer 8

You'll have to expand out the cubic and square terms,
which is going to be a little bit of a pain.

Answer 9

ok well I multiplied it all out and added/crossed out like terms and I was left with this:
\[\frac{ 4h^2+3h^2+2h }{ h}\]

then I factored out an h and was left with this unfactorable expression:
\[4h^2+3h+2\]

Answer 10

apparently the answer is supposed to be 12x^2-6x+2

Answer 11

im not even left with x as a variable im left with h's

Answer 12

what 0_o

Answer 13

Like if you look at the first term,\[\large\rm 4(x+h)^3=4[x^3+3x^2h+3xh^2+h^3]\]Certainly that first term, 4x^3 is going to cancel out with something,
but not the other stuff.

I don't understand how you ended up with no x's >.< hmm

Answer 14

\[\large\rm -3(x+h)^2=-3[x^2+2xh+h^2]\]

Answer 15

Lot of stuff to group together D: oh boy

Answer 16

well I figured that \[-3(x+h)^2 =( -3x^2)+(-3h^2).\]

Answer 17

I see what you did.....and yeah that is going to take me more time to expand

Answer 18

I got to :
\[\frac{ 12x^2h+12x^2+4h^3-6xh-3h^2 }{ h}\]

can I cross out 1 h in every monomial? like this?
\[\frac{ 12x^2h+12x^2+4h^3-6xh-3h^2 }{ h}\] --> \[12x^2+12hx+4h^2-6x-3h\]

Answer 19

I remember definition of the derivative and it is a royal pain... the h must cancel, so factor all terms with the h and cross it out . then whatever h is remaining take the limit. when h = 0. it's almost there, but it's missing a term

the shortcuts (to find the derivative the easy which I don't know why this method isn't taught first) is a great way to check your work
f(x)=4x^3-3x^2+2x-pi

all constants without a variable will derive to 0. so see that pi over there? the derivative of pi is 0 because we treat pi as a constant
f'(x) = 12x^2-6x+2
which is almost correct but somehow the 2 is dropped during the process

Answer 20

whoops take the limit when h approaches 0 X>X

Answer 21

\[\rm f'(x)=\lim_{h\to0}\frac{[4(x+h)^3-3(x+h)^2+2(x+h)-\pi]-\left[4x^3-3x^2+2x-\pi\right]}{h} \]

\[\rm f'(x)=\lim_{h\to0}\frac{[4(x+h)^3-3(x+h)^2+2(x+h)-\pi]+ \left[-4x^3+3x^2-2x+\pi\right]}{h} \]

\[\rm f'(x)=\lim_{h\to0}\frac{[4x^3+12x^2h+12hx^2+4h^3-3x^2-6xh-3h^2+2x+2h-\pi]+ \left[-4x^3+3x^2-2x+\pi\right]}{h} \]

Answer 22

4x^3, 3x^2,2x, and pi should all cancel

Answer 23

\rm f'(x)=\lim_{h\to0}\frac{[12x^2h+12hx^2+4h^3-6xh-3h^2+2h-\pi]}{h}

Answer 24

\[\frac{12x^2h+12h^2x+4h^3-6xh-2h^2+2h}{h}\]

factor all the h's
\[\frac{(h)[12x^2+12hx+4h^2-6x-2h+2]}{h}\]


now the h does cancel and you're left with
\[12x^2+12hx+4h^2-6x-2h+2 \]

take the limit as h approaches 0 for any term with an h left over. those terms should disappear and you should have the derivative. I know I made an error and didn't realize it until the end but I can't edit that. Also, my latex broke and my batteries for my mouse dies.

Answer 25

*died rather.. but on one of the 2's there's an h which goes away when you take the limit but then there's the other 2 so that saves the day

Answer 26

thanks a lot ! I have another quick question. I need to find the derivative for this using the definition :
\[\frac{ x^2 }{ 3}-\frac{ 3 }{ x^2 }\]

When I plug it into the definition is should look like this right? :
\[\frac{ \frac{ (x+h)^2 }{ 3 }-\frac{ 3 }{( x+h)^2} }{ h }\]

Answer 27

@UsukiDoll

Answer 28

I am away from my desktop and need to look at this later. I think it would help if you combine the fraction first and then apply the definition of the derivative

Answer 29

Or maybe not. There would've been x^4 involved and that would involve the binomial theorem for the x+h parts

Answer 30

using the definition of the derivative
\[\frac{f(x+h)-f(x)}{h}\]
we should have
\[\huge\frac{\frac{ (x+h)^2 }{ 3}-\frac{ 3 }{ (x+h)^2 }-(\frac{ x^2 }{ 3}-\frac{ 3 }{ x^2 })}{h}\]

Answer 31

so the gcf is\[ (x+h)^2 (x^2)(3) \]
for the upper fraction... wow this is crazy with the definition version.

Answer 32

sigh all that work for this
\[\LARGE f(x) = \frac{1}{3}x^2-3x^{-2}\]
\[\LARGE f'(x) = \frac{2}{3}x-(-2)3x^{-2-1}\]
\[\LARGE f'(x) = \frac{2}{3}x+6x^{-3}\]
\[\LARGE f'(x) = \frac{2}{3}x+\frac{6}{x^3}\]