Question

find y'

Answer 1

\[y=\frac{ (x ^{2}+5)^{5} }{ (x ^{4}+3)^{3} }\]

Answer 2

well you would do the y=u/v and so then your \[y'=\frac{v(du)-u(dv)}{v^{2}}\]

Answer 3

i know about that but im confused when i simplyfing it

Answer 4

\[y'=\frac{2x(x^2+5)^4(5(x^4+3)-6x^2(x^2+5))}{(x^4+3)^4}\]

Answer 5

A tip, is easier to do it with the product rule
u=(x^2+5)^5
v=1/(x^4+3)^3
Now y'= u'v+v'u

The result is what Mathmuse said.

Answer 6

product rule is may be easier derivcative (no negative sign involved), but cleaning it up later tends to be more labourious

Answer 7

In this case is easier xD