Find dz/dy and dz/dx
Let z = ∫e^(sin(t))dt from x to y
a = x
b = y
I tried thinking about it like a chain rule but even then i'm a little unsure.
I know dz/dt = e^(sin(t)). Can you please point me in the right direction if i'm supposed to use the chain rule.
if F = Integral(f(t)) [x,y] then
dF/dx = f(x) = e^sin(x)
dF/dy = -f(y) = -e^sin(y)
wikipedia has a good article on differentiation under the integral
To find dz/dy and dz/dx, you can use the fundamental theorem of calculus and apply the chain rule. Here's the step-by-step solution:
Step 1: Calculate the integral ∫e^(sin(t))dt from x to y.
Since the bounds of integration are x and y, we will treat y as a constant, and integrate with respect to t. The integral will give us the antiderivative of e^(sin(t)).
Let F(t) be the antiderivative of e^(sin(t)). Hence, ∫e^(sin(t))dt = F(t) + C, where C is the constant of integration.
Step 2: Apply the fundamental theorem of calculus.
According to the fundamental theorem of calculus, if a function G(t) is defined as ∫f(t)dt from a to b, where a and b are constants, then dG(t)/dt = f(t).
In this case, z = ∫e^(sin(t))dt from x to y. So, z = F(t) + C.
Step 3: Apply the chain rule to find dz/dy.
To find dz/dy, we need to differentiate z = F(t) + C with respect to y. Since z does not contain y directly, we can treat y as a constant.
Differentiating both sides of z = F(t) + C with respect to y, we get dz/dy = 0.
Step 4: Apply the chain rule to find dz/dx.
To find dz/dx, we need to differentiate z = F(t) + C with respect to x. Since z does not contain x directly, we can treat x as a constant.
Differentiating both sides of z = F(t) + C with respect to x, we get dz/dx = 0.
Therefore, dz/dy = 0 and dz/dx = 0.