Suppose F(x)=f(x)g(2x) . If f(1)=3 , f′(1)=3 , g(2)=2 , and g′(2)=7 , find F′(1).
F′(1)=
′ is supposed to be the prime(')
F''(x) = f'(x)g(2x) + f(x)g'(2x)
==> F''(1) = (3)(2) + (3)(7) = answer
To find F'(1), we need to use the product rule and chain rule.
The product rule states that if F(x) = f(x)g(x), then F'(x) = f'(x)g(x) + f(x)g'(x).
Applying the product rule to the given function, F(x) = f(x)g(2x), we can rewrite it as:
F(x) = f(x) * g(2x)
Now let's differentiate F(x) using the product rule and chain rule:
F'(x) = f'(x) * g(2x) + f(x) * g'(2x) * (2)
Now we substitute the given values into the equation:
f(1) = 3
f'(1) = 3
g(2) = 2
g'(2) = 7
F'(x) = 3 * g(2x) + f(x) * 7 * 2
Now we need to find F'(1), so we substitute x = 1 into the equation:
F'(1) = 3 * g(2) + f(1) * 7 * 2
Since g(2) = 2 and f(1) = 3,
F'(1) = 3 * 2 + 3 * 7 * 2
Simplifying the expression,
F'(1) = 6 + 42
Therefore, F'(1) = 48.