If y = f(g(x))
dy/dx = dy/df * df/dg * dg/dx
= -csc^2(π/4 u)(π/4) * 6/2√x
= -3πcsc^2(u)/4√x
or, directly,
y=(f◦g)(x) = cot(π/4 * 6√x) = cot(3π√x/2)
y' =
-3πcsc^2(3π√x/2)
----------------------
4√x
find value of (f•g)'(x) at x=1 for f(u)=cot(pi*u/4), u=g(x)=6(square root(x))
By the way, the little dot is supposed to be hollow above, as in f of g of x.
4 answers
Was wondering where the y and X came out of? I thought it was just f and u?
Could you please explain how dg/dx is 6 over 2times square root of x?
Wait sorry, scratch all of that. How did you get df/dg????