Question:

Let f and g be differentiable functions such that f(x) = x^2 - 3x + 1 and g(x) = sin(2x) for all real numbers x. The functions h and k are defined as:

h(x) = f(g(x)) k(x) = g(f(x))

(a) Find an expression for h'(x) in terms of f'(x) and g'(x) using the Chain Rule.

(b) Evaluate h'(π/4) given that f'(x) = 2x - 3 and g'(x) = 2cos(2x).

(c) Find an expression for k'(x) in terms of f'(x) and g'(x) using the Chain Rule.

(d) Evaluate k'(π/3) given that f'(x) = 2x - 3 and g'(x) = 2cos(2x).

Answer:

(a) To find the expression for h'(x) using the Chain Rule, we can consider h as a composition of functions f and g. The Chain Rule states that if h(x) = f(g(x)) is differentiable, then h'(x) = f'(g(x)) * g'(x).

Applying the Chain Rule to our given function h(x) = f(g(x)):

h'(x) = f'(g(x)) * g'(x)

Therefore, h'(x) equals the derivative of f(g(x)) multiplied by the derivative of g(x).

(b) We are given that f'(x) = 2x - 3 and g'(x) = 2cos(2x). To evaluate h'(π/4), we substitute π/4 for x in the expression we found in part (a):

h'(π/4) = f'(g(π/4)) * g'(π/4)

First, we find g(π/4):

g(π/4) = sin(2(π/4)) = sin(π/2) = 1

Next, we substitute g(π/4) = 1 into f'(g(x)) and evaluate it using the given derivative of f(x):

f'(1) = 2(1) - 3 = -1

Finally, we substitute g(π/4) = 1 into g'(x) and evaluate it using the given derivative of g(x):

g'(π/4) = 2cos(2(π/4)) = 2cos(π/2) = 0

Therefore, h'(π/4) = f'(1) * g'(π/4) = (-1) * (0) = 0.

(c) To find the expression for k'(x) using the Chain Rule, we can consider k as a composition of functions g and f. Applying the Chain Rule:

k'(x) = g'(f(x)) * f'(x)

Therefore, k'(x) equals the derivative of g(f(x)) multiplied by the derivative of f(x).

(d) We are given that f'(x) = 2x - 3 and g'(x) = 2cos(2x). To evaluate k'(π/3), we substitute π/3 for x in the expression we found in part (c):

k'(π/3) = g'(f(π/3)) * f'(π/3)

First, we find f(π/3):

f(π/3) = (π/3)^2 - 3(π/3) + 1 = π^2/9 - 3π/3 + 1 = π^2/9 - π + 1

Next, we substitute f(π/3) = π^2/9 - π + 1 into g'(x) and evaluate it using the given derivative of g(x):

g'((π^2/9 - π + 1) = 2cos(2(π^2/9 - π + 1))

Continuing with the calculation:

= 2cos(2π^2/9 - 2π + 2)

Therefore, k'(π/3) = g'((π^2/9 - π + 1)) * f'(π/3) = 2cos(2π^2/9 - 2π + 2) * (2(π/3) - 3).

This gives the expression for k'(π/3) in terms of f'(x) and g'(x).