Here's the question you clicked on:
ksaimouli
find the arc lenght x=(y^3/3)((x^2)+2)^3/2 from x=0 to x=3
\[x=\frac{ y^3 }{ 6 }+\frac{ 1 }{ 2y }\]
\[\frac{ dx}{ dy }=\frac{ y^2 }{ 2 }-\frac{ 1 }{ y^2}\]
\[\int\limits_{1}^{2}\sqrt{1+((\frac{ y^2 }{ 2 })-\frac{ 1 }{ y^2 })^2} dy\]
Woops the 2 should still be there on the second term. A -1 came down from the power rule.
The problem at the very top seems to be different than the one you formatted below it, I'm confused..
when i take derivative using quotient rule i got -1/2y^2
Yes that's the correct derivative. Don't use quotient rule, easier to use power rule.\[\large \left(\frac{1}{2y}\right)'\quad = \quad \left(\frac{1}{2}y^{-1}\right)' \quad = \quad -\frac{1}{2}y^{-2}\]
Do you have 2 separate problems listed? This is rather confusing....