How do I find the derivative of 6 + 8cosx?
-8sinx
To find the derivative of the function 6 + 8cos(x), we can take the derivative term by term using the differentiation rules.
The derivative of a constant term (in this case, 6) is zero. So, the derivative of 6 is 0.
Now, let's find the derivative of the second term, 8cos(x). The derivative of cos(x) is known to be -sin(x). We can apply the chain rule to find the derivative of 8cos(x).
Chain rule states that if we have a function g(f(x)), the derivative of g(f(x)) is g'(f(x)) * f'(x).
In our case, g(f(x)) is 8cos(x), where g(u) = 8u, and f(x) = cos(x).
To find g'(f(x)), we differentiate g(u) = 8u with respect to u, which gives us g'(u) = 8.
To find f'(x), we differentiate f(x) = cos(x) with respect to x, resulting in f'(x) = -sin(x).
Now, applying the chain rule, we have:
Derivative of 8cos(x) = g'(f(x)) * f'(x) = 8 * (-sin(x)) = -8sin(x)
Therefore, the derivative of 6 + 8cos(x) is 0 - 8sin(x), or simply -8sin(x).