find the derivative of the function
integral from cosx to sinx (ln(8+3v)dv)
y'(x)= (sin(8+3((sin(x)))+x)+sin(8+3sin(x)-x))/2
Is this the right answer?????
I gave you my answer which is quite different.
Hmm. I thought that, using Leibniz' Rule,
F'(x) = ln(8+3sinx)*cosx - ln(8+3cosx)(-sinx)
Did I miss something there?
To find the derivative of the given function, we can use the Fundamental Theorem of Calculus. According to the theorem, if a function F(t) is defined as the integral of another function f(t), then the derivative of F(t) is equal to f(t).
In this case, the function F(x) is defined as the integral from cos(x) to sin(x) of ln(8+3v)dv. To find the derivative of F(x), we need to find the derivative of the integrand ln(8+3v) with respect to v, and then evaluate it at the upper limit sin(x) and the lower limit cos(x).
Let's compute the derivative of ln(8+3v) with respect to v first. The derivative of ln(u) with respect to u is 1/u, so we can apply the chain rule here.
d/dv ln(8+3v) = (1)/(8+3v) * d/dv (8+3v) = (1)/(8+3v) * 3
Now, substituting the limits cos(x) and sin(x) into this derivative expression, we get:
d/dx F(x) = [(1)/(8+3(sin(x))) * 3] - [(1)/(8+3(cos(x))) * 3]
Simplifying this expression further,
= 3/(8+3sin(x)) - 3/(8+3cos(x))
Therefore, the derivative of the given function is:
y'(x) = 3/(8+3sin(x)) - 3/(8+3cos(x))
The answer you provided, y'(x) = (sin(8+3((sin(x)))+x)+sin(8+3sin(x)-x))/2, is not correct.