[Ezra Halleck]

22. Find the de­riv­a­tive for a func­tion given as a quo­tient (an­swer must be a sim­ple frac­tion and no neg ex­po­nents!)

[H Pardon]

This was a dif­fi­cult prob­lem!!!
Okay first look at the prob­lem and you will no­tice a neg­a­tive ex­po­nent -3
get rid of the neg­a­tive ex­po­nent by mak­ing it 1/x^3
So ba­si­cally the prob­lem should read xtanx/x2 +1/x^3
the x in the nu­mer­a­tor should can­cel with the de­nom­i­na­tor x^2
tanx/x + 1/x^3
From here ba­si­cally take d/dx (tan/x) + d/dx(1/x^3)

[Devindra]

This is what I got for my an­swer
= tan(x) / ( x^2+ (1/ x^3)) − ((x⋅(2x−3/x^4))⋅tan(x)) / (x^2+(1/x^3))^2 + (x⋅(sec(x))^2 / (x^2+(1/x^3))

[H Pardon]

I got xsec^2x-tanx/x^2 + -3/x^4

[Ezra Halleck]

First step rec­om­mended is to mul­ti­ply nu­mer­a­tor and de­nom­i­na­tor by x^3 to get
x^4 tanx/(x^5+1)
Using quo­tient rule, we get
[5x^8tanx-(4x^3tanx +x^4sec^2x)(x^5+1)]/(x^5+1)^2