If f(x)=e^x, g(x)=sin(x), and
h(x)=f(g(x)), then h'(pi/2)=?
Please help I have no idea how to solve
I just did this a little while ago
h = e^sin x
h' = cos x e^sin x
cos pi/2 = 0
sin pi/2 = 1
so
0 * 1 = 0
To find h'(pi/2), we need to take the derivative of h(x) with respect to x and then evaluate it at x=pi/2.
First, let's find h(x). We are given that h(x) is the composition of f(x) and g(x), so we can write it as:
h(x) = f(g(x))
Substituting the given functions, we have:
h(x) = f(sin(x))
Next, let's find the derivative of h(x) using the chain rule. The chain rule states that if we have a composite function h(x) = f(g(x)), the derivative of h(x) with respect to x is given by:
h'(x) = f'(g(x)) * g'(x)
In this case, f(x) = e^x and g(x) = sin(x). Let's find their derivatives:
f'(x) = d/dx(e^x) = e^x (since the derivative of e^x is e^x)
g'(x) = d/dx(sin(x)) = cos(x) (since the derivative of sin(x) is cos(x))
Now, let's substitute these derivatives into the chain rule equation:
h'(x) = f'(g(x)) * g'(x)
= e^sin(x) * cos(x)
Finally, to find h'(pi/2), substitute x = pi/2 into the equation:
h'(pi/2) = e^sin(pi/2) * cos(pi/2)
Since sin(pi/2) = 1 and cos(pi/2) = 0, we have:
h'(pi/2) = e^1 * 0
= 0
Therefore, h'(pi/2) = 0.