"find the value of the derivative (if it exists) at each indicated extremum. (if an answer does not exist, enter dne.)
f (x) = cos(πx/2)

The derivate is f'(x) = -sin(πx/2)*(π/2)

Step-by-step explanation:

We want the value of the derivate at each point marked:

As you may know, the derivate of cos(x) is equal to -sin(x)

And in our case:

f(x) = cos(πx/2)

f'(x) = -sin(πx/2)*(π/2)

Where the rule used is that the derivate of f(g(x)) is equal to f'(g(x))*g'(x)

Now, as you may see, in both the marked points the curve changes of direction, so the tangent line in that point must be zero, letsinput the values of x in the derivate and see if this is true:

f'(0) = -sin(π*0/2)*(π/2) = -sin(0)*(π/2) = 0

f'(2) = -sin(π*2/2)*(π/2) = -sin(π)*(π/2) = 0

So this makes sense.

here you hope this .

step-by-step explanation:

hypotenuse = √(15² + 8²)

=  √(225 +64)

=  √(289) = 17

sin(x) = cos(y) = 8/17

they are identical so long as you are doing the sine of one angle and the cosine of the only other angle that isn't the right angle.