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.

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

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????

To find the value of (f•g)'(x) at x = 1, we need to take the derivative of the composite function f(g(x)) and then evaluate it at x = 1.

Let's begin by calculating the derivatives of the individual functions.

1. Derivative of f(u) = cot(pi*u/4):
To find the derivative of cot(pi*u/4) with respect to u, we can use the chain rule. The derivative of cot(u) is equal to -csc^2(u). Therefore, the derivative of cot(pi*u/4) is -csc^2(pi*u/4).

2. Derivative of u = g(x) = 6 * sqrt(x):
To find the derivative of g(x), we can apply the power rule. The derivative of sqrt(x) is 1/2 * x^(-1/2), and then the derivative of 6 * sqrt(x) is 6 * (1/2 * x^(-1/2)) = 3 / sqrt(x).

Now, let's find the derivative of the composite function f(g(x)).

Using the chain rule, we have:
(f•g)'(x) = f'(g(x)) * g'(x)

Substituting the derivatives we found earlier:
(f•g)'(x) = (-csc^2(pi*g(x)/4)) * (3 / sqrt(x))

Finally, we can calculate (f•g)'(1) by plugging in x = 1 into the expression we just derived:

(f•g)'(1) = (-csc^2(pi*g(1)/4)) * (3 / sqrt(1))

Since g(1) = 6 * sqrt(1) = 6, we have:
(f•g)'(1) = (-csc^2(pi*6/4)) * 3

You can use a calculator to evaluate the trigonometric function, or simplify further if possible.