Recent content by Michael Santos

What if both functions are to be integrated If both functions are to be integrated what is the argument to integrate?

Homework Statement The indefinite integral $$\int \, $$ and it's argument. The indefinite integral has a function of e.g ## \cos (x^2) \ ## or ## \ e^{tan (x)} \ ## If the argument of ## \cos (x^2) \ ## is ## \ x^2 \ ## then the argument of ## \ e^{tan(x)} \ ## is ## \ x \ ## or ## \ tan (x) \...

Yes i see

Homework Statement Your blowing up a balloon at a rate of 300 cubic inches per minute. When the balloon's radius is 3 inches, how fast is the radius increasing? Homework EquationsThe Attempt at a Solution I know the answer to this question. It is approximately 2.65 inches per minute, what my...

I figured it out thank you.

-1.39...

So $$ \frac {3x^2h+3xh^2+h^3} { \ x^3-5 \ } $$ = h?

How doez that turn into the final answer of $$\frac {3x^2} { \ x^3-5 \ }$$

Well i know it now, what can i do with it?

No i do not It is approaching one

Mark here is what i have so far $$\lim_{h\to 0} {\frac {\ln((x+h)^3 -5) - ln (x^3 - 5)} h} $$ $$\lim_{h\to 0}{\frac {\ln (x^3 +3x^2h +3xh^2 +h^3 -5) -ln (x^3 - 5)} h} $$ $$\lim_{h\to 0}{\frac {\ln (x^3 -5 +3x^2h +3xh^2 +h^3) -ln (x^3-5)} h} $$ I factored and used the product rule $$\lim_{h\to...

Vela here is what i have so far $$\lim_{h\to 0} {\frac {\ln((x+h)^3 -5) - ln (x^3 - 5)} h} $$ $$\lim_{h\to 0}{\frac {\ln (x^3 +3x^2h +3xh^2 +h^3 -5) -ln (x^3 - 5)} h} $$ $$\lim_{h\to 0}{\frac {\ln (x^3 -5 +3x^2h +3xh^2 +h^3) -ln (x^3-5)} h} $$ I factored and used the product rule $$\lim_{h\to...

Yes the basic definition is what i am after. Vela, will you be providing me with the rest of the answer?

How do i move this over to homework

Mentor note: Thread moved from technical math section, so is missing the homework template .[ATTACH=full]234622[/ATTACH]