Question

"Use the definition of the derivative to find the derivative of f(x) = x^2 + 2x at x=5. "

Answer 1

So let's recall what the limit definition of a derivative looks like,\[\large \lim_{h \rightarrow 0} \;\frac{f(x+h)-f(x)}{h}\]

So we see we have an f(x) in the problem, and our f(x) is,\[\large f(x)=x^2+2x\]Our f(x+h) will be,\[\large f(x+\color{royalblue}{h})=(x+\color{royalblue}{h})^2+2(x+\color{royalblue}{h})\]

Now we just plug things in and simplify! :)

Answer 2

Understand how I setup the f(x+h)? Function notation can be a little tricky if you're new to it. Let me know if you're confused on that part.

Answer 3

I am a little new at this, so I do get a little confused on these questions

Answer 4

But I really appreciate that you broke it down for me, thank you =]

Answer 5

It might help to understand function notation if you place brackets around the x's.\[\large f(\color{orangered}{x})=(\color{orangered}{x})^2+2(\color{orangered}{x})\]So the idea is, whatever value you place in the f brackets, it will change every x in the rest of the problem.
\[\large f(\color{salmon}{7})=(\color{salmon}{7})^2+2(\color{salmon}{7})\]\[\large f(\color{royalblue}{x+2})=(\color{royalblue}{x+2})^2+2(\color{royalblue}{x+2})\]


If you need help simplifying it from the definition, let me know. c:

Answer 6

That does make a lot more sense, I normally forget to add brackets, so I occasionally get confused