Find the derivative of the function.
y = integral from cosx to sinx
(ln(8+3v)) dv
lower limit = cosx
upper limit = sinx
y'(x) = ????
Just use Lebniz's Rule:
d/dx ∫[a(x),b(x)] f(t) dt =
f(b(x))b' - f(a(x))a'
d/dx ∫[cosx,sinx] ln(8+3v) dv
= ln(8+3sinx)(cosx) - ln(8+3cosx)(-sinx)
To find the derivative of the function with respect to x, we can use the Fundamental Theorem of Calculus. According to this theorem, if we have a function in the form y = ∫[a(x), b(x)] f(v) dv, where a(x) and b(x) are functions of x, then the derivative of y with respect to x is given by:
y'(x) = f(b(x)) * b'(x) - f(a(x)) * a'(x).
In this case, we have y = ∫[cosx, sinx] ln(8+3v) dv. Let's find the derivative step-by-step:
Step 1: Determine a(x) and b(x)
In our case, a(x) = cosx and b(x) = sinx.
Step 2: Find f(v)
f(v) = ln(8+3v)
Step 3: Find a'(x) and b'(x)
a'(x) = -sinx and b'(x) = cosx
Step 4: Substitute the values into the derivative formula
y'(x) = ln(8+3sinx) * cosx - ln(8+3cosx) * (-sinx)
Simplifying the expression, we have:
y'(x) = cosx ln(8+3sinx) + sinx ln(8+3cosx)
Therefore, the derivative of the given function y = ∫[cosx, sinx] ln(8+3v) dv with respect to x is y'(x) = cosx ln(8+3sinx) + sinx ln(8+3cosx).
To find the derivative of the given integral function, we can use the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus states that if a function f(x) is given as an integral with respect to a variable, say x, then its derivative can be found by evaluating the integrand function at the upper limit of the integral and then subtracting the evaluation of the integrand function at the lower limit. In equation form, it can be written as:
d/dx ∫[a to b] f(v) dv = f(b) - f(a)
Now, let's apply this theorem to the given integral function.
The integral function is:
y = ∫[cos(x) to sin(x)] ln(8+3v) dv
To find y'(x), we need to evaluate the integrand function ln(8+3v) at the upper limit (sin(x)) and subtract the evaluation at the lower limit (cos(x)).
So, the derivative of y with respect to x, y'(x), is given by:
y'(x) = ln(8+3sin(x)) - ln(8+3cos(x))
Therefore, y'(x) = ln(8+3sin(x)) - ln(8+3cos(x)) is the derivative of the given integral function.