Compute the derivative of (f º g).
f(u) = $a·u+1, g(x) = sin($b·x)
Recall the chain rule for differentiating (f º g)(x):
(f º g)'(x) =(f(g(x))' = AJHs0aiOXXYYKYMi58NY5Q==(g(x))·GlwRU+QAspKTpwBOi8WbrQ==
For the given functions, observe that f is a function of u, and g is a function of pdlFLKGakEU=.
In order to substitute appropriately into the chain rule to compute,
(f º g)'(x) = (f(g(x)))' = f'(g(x))g'(x),
we need to find f'(g(x)), which is f'(u) evaluated at g(x) = sin($b·x).
f(u) = $a·u+1
f'(u) = nc1ItEz0kR4=
Thus, for all u = g(x) = sin($b·x),
f'(g(x)) = f'(sin($b·x)) = nc1ItEz0kR4=.
Next we calculate g'(x).
g(x) = sin($b·x)
g'(x) = iSba6t70dtA=·cos(iSba6t70dtA=·x)
Applying the chain rule, find the derivative of (f º g) where
f(u) = $a·u+1 and g(x) = sin($b·x)
with
f'(g(x)) = $a and g'(x) = $b·cos($b·x).
(f º g)'(x) = f'(g(x))g'(x) = SFgqQUkJGdg=·cos(iSba6t70dtA=·x)