Use Leibniz's rule to find dy/dx.
**Leibniz's rule: If g(x) and h(x) are differentiable functions and f(u) is continuous for u between g(x) and h(x), then
d/dx of the integral from g(x) to h(x) of f(u) du = f[h(x)]h'(x) - f[g(x)]g'(x)
y = the integral from 2+x² to 2 of (cot t) dt
What's to worry about? You have the formula, just plug in f,g,h. The only possible sticking spot is knowing that Int(cot(t) dt] = ln sin t
To find dy/dx using Leibniz's rule, we need to determine the derivative of the integral with respect to x. In this case, we have:
y = ∫[2+x² to 2] cot(t) dt
To apply Leibniz's rule, we need to identify the functions g(x), h(x), and f(u). Here:
g(x) = 2 + x²
h(x) = 2
f(u) = cot(u)
Now, let's calculate dy/dx using Leibniz's rule formula:
dy/dx = f[h(x)] * h'(x) - f[g(x)] * g'(x)
First, let's find f[h(x)]:
f[h(x)] = cot(h(x))
= cot(2)
Next, let's find h'(x):
h(x) = 2
h'(x) = 0
Now, let's find f[g(x)]:
f[g(x)] = cot(g(x))
= cot(2 + x²)
Finally, let's find g'(x):
g(x) = 2 + x²
g'(x) = 2x
Putting it all together, we have:
dy/dx = cot(2) * 0 - cot(2 + x²) * 2x
Thus, the derivative of y with respect to x (dy/dx) using Leibniz's rule is cot(2 + x²) * (-2x).