How do I find the derivative of
⌠lnx
⌡π cos(e^(t))dt
The question above is asking for the derivative of the integral cos(e^(t))
from pi to lnx
since
d/dx ∫[a(x),b(x)] f(t) dt = f(b(x))db/dx - f(a(x))da/dx, we have
cos(e^(lnx)) d/dx(ln(x)) - cos(e^π) d/dx(π)
= cos(x) * 1/x - 0
= cos(x)/x
check the examples on wikipedia for differentiation under intagral.
okay thank you!
To find the derivative of the given expression, we can use the Fundamental Theorem of Calculus combined with the Chain Rule for differentiation.
The Fundamental Theorem of Calculus states that if we have an integral of the form
∫[a to b] f(t) dt
and F(t) is an antiderivative of f(t), then the derivative of the integral with respect to t is equal to the integrand evaluated at the upper limit multiplied by the derivative of the upper limit, i.e.,
d/dx ∫[a to b] f(t) dt = f(b) * d/dx b
In this case, we have the integral
∫[π to lnx] cos(e^(t)) dt
To find the derivative with respect to x, we first evaluate the integral using the Fundamental Theorem of Calculus.
Since cos(e^(t)) is the integrand, we need to find an antiderivative of cos(e^(t)). The derivative of e^(t) is e^(t), so we have to take the chain rule into account. The antiderivative of cos(e^(t)) is sin(e^(t)), and we'll use this to evaluate the integral.
∫[π to lnx] cos(e^(t)) dt = sin(e^(lnx)) - sin(e^(π))
Now that we have the result of the integral, we can differentiate it with respect to x.
Using the Chain Rule, we find that the derivative of sin(e^(lnx)) is cos(e^(lnx)) * d/dx(e^(lnx)), and the derivative of sin(e^(π)) is 0 (since it's a constant).
The derivative of e^(lnx) with respect to x is obtained by the Chain Rule, which says that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). In this case, f(u) = e^u and g(x) = lnx, so we have u = lnx, f'(u) = e^u, and g'(x) = 1/x.
Therefore, the derivative of e^(lnx) with respect to x is e^(lnx) * (1/x) = x * (1/x) = 1.
Putting it all together, we have:
d/dx ∫[π to lnx] cos(e^(t)) dt = cos(e^(lnx)) * 1 - 0 = cos(e^(lnx))
So, the derivative of the given expression is cos(e^(lnx)).