evaluate the table and find d/dx [g[f(2x)]] at x=1
f(x) 6,1,6,2
f'(x) 1,2,5,7
g(x) 1,4,4,3
g'(x) 4,5,5,-4
Assuming the x values for the table are 1,2,3,4 then
d/dx g(f(2x))
= g'(f(2x)) * f'(2x) * 2
= g'(f(2)) * f'(2) * 2
= g'(1) * 2 * 2
= 5*2*2
= 20
consider something like y = e^sin(2x)
y'(x) = e^sin(2x) * cos(2x) * 2
To find d/dx [g[f(2x)]] at x=1, we will follow these steps:
Step 1: Substitute 2x into f(x) since we have f(2x) in the expression. We get f(2x) = 6, 1, 6, 2.
Step 2: Calculate g[f(2x)] by substituting each value of f(2x) into g(x). g[f(2x)] = g(6), g(1), g(6), g(2). Evaluating these values with the given table, we get g[f(2x)] = 4, 4, 4, 4.
Step 3: Now we have g[f(2x)] = 4, 4, 4, 4. We need to find d/dx [g[f(2x)]] at x=1.
Step 4: To find d/dx [g[f(2x)]] at x=1, we will differentiate g[f(2x)] with respect to x.
Using the chain rule, the derivative of g[f(2x)] is g'[f(2x)] * f'(2x).
Step 5: Substitute each value of f(2x) and g'[f(2x)] into the expression g'[f(2x)] * f'(2x).
We get:
At f(2x) = 6, g'[f(2x)] = g'(6) = 5, and f'(2x) = f'(6) = 7.
At f(2x) = 1, g'[f(2x)] = g'(1) = 5, and f'(2x) = f'(2) = 2.
At f(2x) = 6, g'[f(2x)] = g'(6) = 5, and f'(2x) = f'(6) = 7.
At f(2x) = 2, g'[f(2x)] = g'(2) = 5, and f'(2x) = f'(4) = 5.
Therefore, d/dx [g[f(2x)]] at x=1 is (5 * 7) when f(2x) = 6, (5 * 2) when f(2x) = 1, (5 * 7) when f(2x) = 6, and (5 * 5) when f(2x) = 2.
Simplifying, we find:
d/dx [g[f(2x)]] at x=1 is 35 when f(2x) = 6, 10 when f(2x) = 1, 35 when f(2x) = 6, and 25 when f(2x) = 2.