Find derivative of y = x^2 + (3/x) using formal definition

[Miszka]

Hey. I can't manage to do one task that I have to do at home. It's pretty simple: "Use the formal definition of the derivative and the rules for limits to find the derivatives of the functions".

The formal definition in my book is [MATH]f'(x)=\lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}[/MATH]
I have to find the derivative of that guy: [MATH]x^{2} + \frac{3}{x}[/MATH]
It seems easy but I just have no idea how to do that, I've tried many times. Could you please help me? I'd like to know how to do that thing. Thank you so much in advance!

 

[pka]

"Use the formal definition of the derivative and the rules for limits to find the derivatives of the functions". The formal definition in my book is [MATH]f'(x)=\lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}[/MATH]I have to find the derivative of that guy: [MATH]x^{2} + \frac{3}{x}[/MATH]

\(\displaystyle f(x)=x^2+\dfrac{3}{x}\) then \(\displaystyle f(x+\Delta x)=(x+\Delta x)^2+\dfrac{3}{x+\Delta x}=(x^2+2x\Delta x+(\Delta x)^2)+\dfrac{3}{x+\Delta x}\)
Can you finish. JUST SIMPLIFY.

 

[HallsofIvy]

The only "hard part" (and its not all that hard!) is \(\displaystyle \frac{3}{x+ \Delta x}- \frac{3}{x}\). Getting the common denominator, \(\displaystyle \frac{3x}{x(x+ \Delta x)}- \frac{3(x+ \Delta x)}{x(x+ \Delta x)}= \frac{3x- 3x- \Delta x}{x(x+ \Delta x)}= \frac{-\Delta x}{x^2+ x\Delta x}\). What happens when you divide that by \(\displaystyle \Delta x\)?