Use the table below to evaluate the d/dx[g(f(3x))] at x = 1.
x 1 2 3 4
f(x) 6 1 2 2
f ′(x) 6 1 10 2
g(x) 1 4 4 3
g ′(x) 4 5 7 –4
My work:
D/Dx(g(f(3x)))= g'(f(3x)) * f'(3x) * 2
D/dx (g(f(3*1))))
g'(f(3*1))*f'(3*1) *2
4*3*2=24
I know 24 isn't the right answer
To evaluate d/dx[g(f(3x))] at x = 1, we need to apply the chain rule. The chain rule states that if we have a composite function, the derivative is given by the derivative of the outer function times the derivative of the inner function.
Let's break down the problem step by step:
Step 1: Evaluate f(3x) at x = 1
For x = 1, we have f(3x) = f(3*1) = f(3) = 2.
Step 2: Evaluate g(f(3x)) at x = 1
Using the result from Step 1, we have g(f(3x)) = g(2). Therefore, we need to find g(2).
Looking at the table, when x = 2, g(x) = 4. So, g(2) = 4.
Step 3: Evaluate f'(3x) at x = 1
For x = 1, we have f'(3x) = f'(3*1) = f'(3) = 10.
Step 4: Evaluate g'(f(3x)) at x = 1
Using the result from Steps 1 and 2, we have g'(f(3x)) = g'(2). We need to find g'(2).
Looking at the table, when x = 2, g'(x) = 5. So, g'(2) = 5.
Step 5: Combine all the results to find the final answer
Using the chain rule, we have d/dx[g(f(3x))] = g'(f(3x)) * f'(3x).
Plugging in the values we found:
d/dx[g(f(3x))] = g'(f(3x)) * f'(3x) = 5 * 10 = 50.
Therefore, d/dx[g(f(3x))] at x = 1 is equal to 50.
Work from inside the parentheses,
(g(f(3x))' for x=1
=(g(f(3(1)))'
=(g(f(3))'
=(g(2))' [from table]
=g'(2)
=5 [from table]