Derivatives of quadratic equations :)

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\[f(x)=ax^2+bx+c \\ \text{ so you want to differentiate } f\] First question: Are you doing short cuts or the definition way?
I'm not sure, I just would like to learn how to do this so I'm not always confused lol
You are probably doing definition I bet.

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\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \text{ or } f'(x)=\lim_{z \rightarrow x} \frac{f(z)-f(x)}{z-x}\] you can use either one of these these are basically the same thing they are the same thing it can be shown with a substitution Let's play with the first one...
\[f(x)=ax^2+bx+c \\ f(x+h)=a(x+h)^2+b(x+h)+c\] To find f(x+h) notice I just replace all the x's in the first expression with (x+h)'s.
the numerator in my definition is f(x+h)-f(x) so I could go ahead and look at simplifying that first I will state what is without the simplification : \[f(x+h)-f(x)=[a(x+h)^2+b(x+h)+c]-[ax^2+bx+c]\] The next part is to expand and cancel.
Can you expand (x+h)^2
\[x^2+2hx+h^2\]
cool so I'm going to put that in place of (x+h)^2 ... \[f(x+h)-f(x)=[a(x^2+2xh+h^2)+b(x+h)+c]-[ax^2+bx+c] \\ \text{ now distributing a bit } \\ f(x+h)-f(x)=ax^2+2axh+ah^2+bx+bh+c-ax^2-bx-c\] do you have a problem with my distributing step?
Nope, easy peasy
you should see some things that cancel (or zero out) like ax^2-ax^2=0 and bx-bx=0 and c-c=0
\[f(x+h)-f(x)=\cancel{ax^2}+2axh+ah^2+\cancel{bx}+bh+\cancel{c}-\cancel{ax^2}-\cancel{bx}-\cancel{c}\]
So we left with \[f(x+h)-f(x)=2axh+ah^2+bh\]
right!
now let's go back to our definition this was only the numerator
Okay :)
\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{2axh+ah^2+bh}{h}\]
so... notice that every term on top has a common factor of h and the denominator also has a common factor h
\[f'(x)=\lim_{h \rightarrow 0} \frac{h(2ax+ah+b)}{h(1)}\]
Oh!
we can write as: \[f'(x)=\lim_{h \rightarrow 0} \frac{h}{h} \cdot \frac{2ax+ah+b}{1}\]
which the h's cancel, right?
yep and that whole h on the bottom was the thing that was giving us problems but guess what?
it is no longer there because we canceled that mean h guy on bottom
:)
so we can plug in 0 now
for h that is
and it leaves us with 2ax+b?
\[f'(x) =\lim_{h \rightarrow 0} (2ax+ah+b)=2ax+a(0)+b=2ax+0+b=2ax+b\] you are right
Is that it?
\[f(x)=ax^2+bx+c \text{ gives us } f'(x)=2ax+b\]
yes now you can find the derivative of any quadratic
you actually have a formula for it now we just found a formula for it above
Awesome! Didn't know it was that simple...the video I watched on it made it seem so confusing lol
So there is never a c...
\[f(x)=ax^2+bx+c \text{ gives } f'(x)=2ax+b \\ \text{ example } \\ \text{ say the question is find } f' \text{ given } f(x)=5x^2+3x-4 \\ \text{ we only need to know } a \text{ and } b \\ f'(x)=2(5)x+3=10x+3\]
Well the derivative of a constant will always be zero.
Schnazzy!
Because the slope of y=a constant is zero.
Thank you so very much! :)
Np. So this is exactly what you wanted right? Do you have anymore questions about differentiating a quadratic?
Yes and no, I don't! :)
Wait...perhaps one question...this works for all quadratic equations?
Alright. I might do one more thing before I leave... A different strategy... Instead of expanding...grouping and factoring... \[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{ a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{a[(x+h)^2-x^2]+b((x+h)-x]+[c-c]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{a[(x+h)^2-x^2]+b[(x+h)-x]+0}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{a[(x+h-x)(x+h+x)+b[x+h-x]}{h} \\ \\ f'(x)=\lim_{h \rightarrow 0} \frac{ah(2x+h)+bh }{h} \\ f'(x)=\lim_{h \rightarrow 0} a(2x+h)+b \\ f'(x)=a(2x+0)+b =2ax+b\]
and yes to your question you do know a quadratic function is in this form right: \[f(x)=ax^2+bx+c ?\]
yes!
and if you do then yes to your question
I love quadratic equations....and finding the solutions using the quadratic formula too :)
do you know that you can find the x-coordinate of vertex of the parabola using the derivative
I did not!
Yep. You will probably learn this later. But You can use the derivative to either find the max or min of a function. You can set the derivative equal to zero and find where it does not exist...These will be critical numbers... So you have \[f(x)=ax^2+bx+c \\ f'(x)=2ax+b \\ f'(x)=0 \text{ gives you the equation } 2ax+b=0 \\ \\ \text{ solving for } x \text{ gives you } x=\frac{-b}{2a}\]
This is exactly what we would get if we did the whole completing the square thing.
\[f(x)=ax^2+bx+c \\ f(x)=ax^2+\frac{a}{a}bx+c \\ f(x)=a(x^2+\frac{1}{a}bx)+c \\ f(x)=a(x^2+\frac{b}{a}x)+c \\ f(x)=a(x^2+\frac{b}{a}c+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ f(x)=(x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2 \\ \text{ where the vertex is } (-\frac{b}{2a}, c-a(\frac{b}{2a})^2)\]
But anyways all of that is extra.
Thank you again :)
np

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