Question

F'(x) =(3cos 4x - 28x² +6,7x - 40,8)'

Answer 1

$f'(x)=(3cos4x - 28x^2 +6,7x - 40,8)' \\ (3cos4x)'=-12sin4x \\ (28x^2)'=56x \\ (6,7x)=6,7 \\ (40,8)'=0 \\ f'(x)=-12sin4x-56x+6,7$

Answer 2

To find the derivative of the given function $F(x) = 3\cos(4x) - 28x^2 + 6.7x - 40.8$, you can differentiate each term separately.

1. The derivative of $3\cos(4x)$ with respect to $x$ can be found using the chain rule. The derivative of $\cos(4x)$ is $-4\sin(4x)$, and then you multiply by the derivative of the inner function, which is 4. So, it becomes $-4 \cdot 4\sin(4x) = -16\sin(4x$.

2. The derivative of $-28x^2$ with respect to $x$ is $-56x$.

3. The derivative of $6.7x$ is simply $6.7$.

4. The derivative of $-40.8$ is 0 because it's a constant.

Now, let's combine these derivatives to find $F'(x)$:

$F'(x) = -16\sin(4x) - 56x + 6.7$

So, the derivative of $F(x)$ is $-16\sin(4x) - 56x + 6.7$.