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Mathematics 17 Online
OpenStudy (anonymous):

use the definition of the derivative to find f'(x) ; f(x)=-5x

OpenStudy (zepp):

Do you know what is the definition of the derivative?

OpenStudy (anonymous):

yes. lim f(x+deltax)-f(x) / delta x as delta x approaches 0

OpenStudy (zepp):

Excellent, can you try to plug the function into that formula, and solve?

OpenStudy (anonymous):

uhh when i plugged it in i got 0. but the back of the book says -5

OpenStudy (zepp):

The answer is indeed -5

OpenStudy (zepp):

Okay, let's go with a fact to make this all ezpz :D\[\large \textbf{FACT:} \\ \large \frac{d}{dx}f(x)=\lim _{\Delta x \rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\\large \frac{d}{dx}cf(x) = c\frac{d}{dx}f(x)=c*\lim _{\Delta x \rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x}, \textrm{Where c is a constant.}\]

OpenStudy (zepp):

Let's use this fact to solve our problem: \[\large \frac{d}{dx}f(x)=\frac{d}{dx}(-5x)\]Since -5 is just a constant, we can simply factor it out of the derivative: \[\large -5(\frac{d}{dx}x)\]

OpenStudy (anonymous):

umm what's d?

OpenStudy (zepp):

\(\LARGE \frac{d}{dx}\) is a just a notation to say 'Derivative of'.

OpenStudy (zepp):

With respect to x.

OpenStudy (zepp):

So, \[\large -5(\frac{d}{dx}x)=-5*\lim _{\Delta x \rightarrow0}\frac{x+\Delta x-x}{\Delta x}\]Can you go on and solve it now?

OpenStudy (anonymous):

you plug -5 into x right??

OpenStudy (zepp):

-5 is a constant and it's factored out of the derivative, so you only have to solve the limit now.

OpenStudy (anonymous):

oh then it's -5

OpenStudy (zepp):

The result to the limit is simply 1, 1*-5 = -5

OpenStudy (anonymous):

wait how did you factor it out of the derivative

OpenStudy (zepp):

Because a constant does not affect the limiting process.

OpenStudy (anonymous):

ohhhhh

OpenStudy (zepp):

yep yep yep :)

OpenStudy (anonymous):

how come when i plug -5x in the equation it ends up as 0??

OpenStudy (zepp):

Let's see: \[\large \begin{align} &\frac{d}{dx}(-5x)\\ &=\lim _{\Delta x \rightarrow0}\frac{-5(x+\Delta x)-(-5x)}{\Delta x}\\ &=\lim _{\Delta x \rightarrow0}\frac{-5x-5\Delta x+5x}{\Delta x}\\ &=\lim _{\Delta x \rightarrow0}\frac{-5\Delta x}{\Delta x}\\ &=\lim _{\Delta x \rightarrow0}-5\\ &\textbf{The limit is now just useless, take it off.}\\ &=-5 \end{align}\]

OpenStudy (anonymous):

wait why did you put -5 in front of (x+delta x)

OpenStudy (zepp):

Because I put -5x all together in the limit, and replace the x in this function by x+ delta x.

OpenStudy (anonymous):

wait what do you mean by that

OpenStudy (zepp):

The notation f(x+delta x) mean that you'll need to replace the x in the function by the value in the parentheses, x+delta x, therefore, if you have -5x, f(delta x+ x) would be -5(delta x+ x)

OpenStudy (anonymous):

WTF I DID NOT KNOW THAT UGHHHHHH OKAY

OpenStudy (anonymous):

OK THANK YOU!!!!

OpenStudy (zepp):

Alright, no problem :)

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