Respuesta :
[tex]y = e^{-sin(x)}[/tex]
[tex]y' = -e^{-sin(x)}(cos(x))[/tex]
[tex]0 = -e^{-sin(x)}(cos(x))[/tex]
When is cos (x) = 0? at 90° (aka π/2)
Answer: [tex]\frac{\pi}{2}[/tex]
Answer: Hello there!
We have the function f(x) = exp(-sin(x)) and we want to see in wich point the slope is 0.
This is equivalent to see when f'(x) = 0
where f'(x) is the derivative of f(x)
Then the first step is derivate the function f(x)
if we have a function of the form g(h(x)), his derivate is:
g'(h(x))*h'(x)
In our case, g(x) = exp(x) and h(x) = - sin(x)
then f'(x) = exp(-sin(x))*(-cos(x))
Now we know that exp(-sin(x)) is never equal to 0, then we need to se when the cosine is equal to zero.
cos(90°) = 0, and 90° is equivalent to pi/2
Then f'(pi/2) = exp(-sin(pi/2))*(-cos(pi/2)) = 0
this means that the function f(x) has a slope of 0 in the point x = pi/2