Find the derivative

Question

Suppose F(x)=f(x)g(2x).
If f(1)=3, f′(1)=3, g(2)=2, and g′(2)=5, find F′(1).

 

Derivative of g(2x) at x=1 is?

Answer 1

F'(x)=f'(x)*g(2x)+f(x)*g'(2x)*2

F'(1)=f'(1)*g(2)+f(1)*g'(2)*2

F'(1)=3*2+3*5*2

F'(1)=6+30

F'(1)=36

Answer 2

Note that the derivative of g(2x) with respect to x is: 2 * g'(2x).

 

From the chain rule, we have:

 

F'(x) = f'(x)*g(2x) + f(x) * (2 * g'(2x))

 

Therefore: F'(1) = f'(1) * g(2) + f(1) * (2 * g'(2)) = 3 * 2 + 3 * (2 * 5) = 6 + 3 * 10 = 36.

Answer 3

F'(x) = f'(x) * g(2x) + 2* f(x) * g'(2x)

=> at x = 1, F'(1) = f'(1) * g(2) + 2 * f(1) * g'(2) = 3*2 + 2*3*5 = 6 + 30 = 36

 

derivative of g(2x) = d/dx [g(2x)] = g'(2x) * d/dx [2x] = g'(2x) * 2 = 2 * g'(2x)

=> at x =1, derivative of g(2x) = 2 * g'(2) = 2 * 5 = 10

Answer 4

F`(1)=f`(1)g(2)+f(1)g`(2)=(3)(2)+(3)(5)=21