find the derivative of the function
integral from cosx to sinx (ln(8+3v)dv)
the derivative of a definite integral is how much the area changes as you change the limits.
d/dx Int from p(x) to q(x) of f(v)dv
= [f(q)-f(p)]dx
(draw a sketch of that graph)
sorry, extra dx in there
= f(q) - f(p)
f(q) = ln (8+3sin x)
f(p) = ln (8+3cos x)
f(q)-f(p) = ln [ (8+3sin x)/(8+3cos x) ]
y'(x)= (sin(8+3((sin(x)))+x)+sin(8+3sin(x)-x))/2
Is this the right answer?????
Well, I gave you my answer which is quite different.
To find the derivative of the given function, we can apply the Fundamental Theorem of Calculus.
Let's denote the function as F(x):
F(x) = ∫[cos(x) to sin(x)] ln(8+3v) dv
First, we need to find the antiderivative of ln(8+3v). To do this, we can use u-substitution.
Let u = 8 + 3v
Then, du/dv = 3
Rearranging the equation, dv = du/3
Now, we substitute back into the original integral:
F(x) = ∫[cos(x) to sin(x)] ln(u) * (du/3)
To simplify the integral, we can take out the constant (1/3) outside the integral:
F(x) = (1/3) * ∫[cos(x) to sin(x)] ln(u) du
Integrating ln(u) with respect to u gives us u * ln(u) - u:
F(x) = (1/3) * [u * ln(u) - u] evaluated from cos(x) to sin(x)
Now, substitute u back in terms of v:
F(x) = (1/3) * [(8 + 3v) * ln(8 + 3v) - (8 + 3v)] evaluated from cos(x) to sin(x)
Next, we need to evaluate the function at the upper limit (sin(x)) and subtract the evaluation at the lower limit (cos(x)):
F(x) = (1/3) * [(8 + 3sin(x)) * ln(8 + 3sin(x)) - (8 + 3sin(x))] - [(8 + 3cos(x)) * ln(8 + 3cos(x)) - (8 + 3cos(x))]
Finally, we differentiate F(x) with respect to x to find the derivative:
F'(x) = d/dx [(1/3) * [(8 + 3sin(x)) * ln(8 + 3sin(x)) - (8 + 3sin(x))] - [(8 + 3cos(x)) * ln(8 + 3cos(x)) - (8 + 3cos(x))]]
Now, we take the derivative term by term, using the chain rule and product rule where necessary.
I hope this helps! Let me know if you have any further questions.