I just spent 4 hours finishing a 14 problem set of homework. I’m so annoyed. It’s all because I kept getting confused about what I was being asked, so if you are feeling a bit confused, I am about to lay it all out for myself and whoever may benefit from this.

Derivative = Instantaneous Rate of Change = Slope of the Tangent Line = Limit of the Difference Quotient.

Difference Quotient = Slope of Secant Line = Average Rate of Change

So if given a function, such as f(x) = 2x^2 + 3 and asked to find the slope of the tangent line, all you have to do is find the limit of the difference quotient. For example:

\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h} must be solved for first \dfrac{2(x+h)^2+3-(2x^2+3)}{h} \dfrac{2x^2+4xh+h^2+3-(2x^2+3)}{h}=\dfrac{4xh+2h^2}{h} this can be simplified to 4x +2h now you have to solve for the limit.

4x +2(0) = 4x f'(x)=4x which is also your tangent line slope.

Don’t make the mistake I made, where they asked me to find the secant line’s slope and I kept giving the stupid TANGENT line slope. I am so irritated at myself about this. 4 hours, people… 4.  It’s just ridiculous.

Thanks to komplexify.com for the image

If they ask you to solve for the slope of the secant line all you have to do is give the difference quotient which was (in the equation above) 4x+2h

The other issue I know a lot of people (including myself) have been struggling with is with square roots and the difference quotient.

So I’m going to go ahead and post how to solve for the square root as simply but detailed as I can.

So let’s say you have the function f(x) = \sqrt4x. And you have to solve for the derivative, which cannot have a zero on bottom, and when you make h=0, and you get 0 as your denominator (which is inevitable with this equation)… how do you work around this? Well, you have to rationalize the numerator. Rationalize the what now? I’ll show you…

First let’s remember our difference quotient: \dfrac{f(x+h)-f(x)}{h}

Now we’ll plug in our stuff: \dfrac{\sqrt{4(x+h)}-\sqrt{4x}}{h}

How do we rationalize it? We have to multiply it by itself, just like rationalizing the denominator.. sorta. It should look like so:

\dfrac{\sqrt{4(x+h)}-\sqrt{4x}}{h} * \dfrac{\sqrt{4(x+h)}+\sqrt{4x}}{\sqrt{4(x+h)}+\sqrt{4x}}

It looks confusing but what I like to do from here is show you exactly how you would multiply the top guys to get an easy number:

\sqrt{4(x+h)}-\sqrt{4x} * \sqrt{4(x+h)}+\sqrt{4x} = 4x+4h - 4x - \sqrt{4x} * \sqrt{4(x+h)}+\sqrt{4x} * \sqrt{4(x+h)}

which could be simplified into 4x+4h – 4x = 4h and then put it back on top of the denominator and it now looks like this:

\dfrac{4h}{h\sqrt{4(x+h)}+\sqrt{4x}} cancel out the h’s and you have your new solution

\dfrac{4}{\sqrt{4(x+h)}+\sqrt{4x}}

Now that it’s simplified if you need to find the Derivative/Slope of the Tangent Line/Instantaneous Rate of Change you can take the limit of the difference quotient:

lim_{h\to0}\dfrac{4}{\sqrt{4(x+0)}+\sqrt{4x}} and you get f'(x)=\dfrac{4}{2\sqrt{4x}} aka \dfrac{2}{\sqrt{4x}}

My brain hurts.