calculus
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.
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Answers
First, separate the integral into two parts, dy and dx, that can be differentiated: z(y) = ∫e^(sin(t))dt from x to y z(x) = ∫e^(sin(t)) dt from x to y Then you can use the chain rule to differentiate each part separately. For dz/dy: dz/dy = dz/dt•dt/dy = e^(sin(t))•1 = e^(sin(t)) For dz/dx: dz/dx = dz/dt•dt/dx = e^(sin(t))•(-1) = -e^(sin(t))
