last one

Question

last one

idku · Answer

$$y'=\int\limits_{1-3x}^{1}\frac{u^3}{1+u^2}~du$$going by the fundimental theorem,
$$f(x)=\int\limits_{g(x)}^{C} s(u)~du$$$$f'(g(x))=-s(g(x))~\times g'(x)$$

idku · Answer

In this case,
$$f'(u)=~-\frac{u^3}{1+u^2}\times (\frac{d}{dx}~u)$$$$f'(1-3x)=~-\frac{(1-3x)^3}{1+(1-3x)^2}\times (\frac{d}{dx}~1-3x)$$$$f'(1-3x)=~-\frac{(1-3x)^3}{1+(1-3x)^2}\times (-3)$$$$f'(1-3x)=~\frac{3(1-3x)^3}{1+3(1-3x)^2}$$

idku · Answer

On correct, the very first line should say y=...

idku · Answer

(not y')

idku · Answer

"on correct" was supposed to be "one correction"

idku · Answer

oh I wrote d/dx of u, should be d/du of u

myininaya · Answer

i think i can't complain

idku · Answer

Very nice, I got some brain!

idku · Answer

So, will take it as correct. Ty

myininaya · Answer

You got a lot of brain.

idku · Answer

Not in math, I really would never see my self in math. I would, if I was on a lower than calc1 level. But this is something I should know for sure, and I don't if I need help with it.

anyway, (I guess) tnx for helping me once again!
(I could at least notice that it was negative as a lower limit of 1-3x   that is nice)

idku · Answer

Done for now...

myininaya · Answer

you did really good
and yeah i think you are done for now

idku · Answer

yeah got basic integrals should be able to do