Use part 1 of the Fundamental Theorem of Calculus to find the derivative of
g(x)= integrate from 9x to 4x of ((u+4)/(u-1))du
To find the derivative of the given function g(x), which involves an integral, we can use part 1 of the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus states that if F(x) is the integral of f(t) with respect to t from a to x, then the derivative of F(x) with respect to x is equal to the function f(x).
Now let's apply this theorem to the function g(x):
g(x) = ∫(9x to 4x) ((u+4)/(u-1)) du
To evaluate this integral, we will use u-substitution. Let u = 4x, then du = 4dx. We also need to change the limits of integration accordingly.
When x = 9x, u = 4(9x) = 36x
When x = 4x, u = 4(4x) = 16x
Now, we can rewrite the integral in terms of u:
g(x) = ∫(36x to 16x) ((u+4)/(u-1)) du
Next, we need to find the antiderivative of ((u+4)/(u-1)) with respect to u. Let's call this antiderivative function F(u).
F(u) = ∫((u+4)/(u-1)) du
To find F(u), we can use partial fraction decomposition:
((u+4)/(u-1)) = A + (B/(u-1))
Multiplying through the denominators and equating the numerators, we have:
u + 4 = A(u-1) + B
Expanding and simplifying the right side, we get:
u + 4 = Au - A + B
By comparing coefficients, we find A = 1 and -A + B = 4, which gives B = 5.
Therefore, ((u+4)/(u-1)) = 1/(u-1) + 5/(u-1).
Now, we can rewrite F(u) using the partial fraction decomposition:
F(u) = ∫(1/(u-1) + 5/(u-1)) du
Split the integral:
F(u) = ∫(1/(u-1)) du + ∫(5/(u-1)) du
Integrating each term, we get:
F(u) = ln|u-1| + 5ln|u-1| + C
Combining the terms and factoring out ln|u-1|, we have:
F(u) = 6ln|u-1| + C
Now, let's substitute u = 36x and u = 16x back into F(u):
F(u) = 6ln|36x-1| + C
Finally, we can find the derivative of g(x) by taking the derivative of F(u) with respect to x:
dg(x)/dx = d(6ln|36x-1| + C)/dx
Since C is a constant, its derivative is zero, so we can ignore it:
dg(x)/dx = d(6ln|36x-1|)/dx
Using the chain rule, the derivative of ln|36x-1| with respect to x is:
dg(x)/dx = 6(1/(36x-1))(36)
Simplifying the expression, we have:
dg(x)/dx = 6/(36x-1)
Therefore, the derivative of g(x) is 6/(36x-1).
To find the derivative of the given function using the Fundamental Theorem of Calculus, we need to use Part 1 of the theorem. Part 1 states that if a function g(x) is defined as the integral of a function f(u) with respect to u over an interval from a constant lower limit to x, then the derivative of g(x) with respect to x is equal to f(x).
In this case, we have the function g(x) defined as the integral of ((u+4)/(u-1)) du over the interval from 9x to 4x. To find the derivative of g(x), we need to evaluate the integrand at x and then differentiate it with respect to x.
Let's break down the steps to find the derivative:
Step 1: Evaluate the integrand at x:
Replace u in the integrand ((u+4)/(u-1)) du with x to get ((x+4)/(x-1)).
Step 2: Differentiate the integrand with respect to x:
Differentiate the expression ((x+4)/(x-1)) with respect to x.
To differentiate ((x+4)/(x-1)), we will use the quotient rule. The quotient rule states that if we have a function h(x) defined as g(x)/f(x), then the derivative of h(x) is given by [f(x)g'(x) - g(x)f'(x)] / [f(x)]^2.
Applying the quotient rule, we have:
g(x) = (x+4) and f(x) = (x-1)
g'(x) = 1 (derivative of x+4 with respect to x)
f'(x) = 1 (derivative of x-1 with respect to x)
Let's plug these values into the quotient rule formula:
[g'(x)f(x) - g(x)f'(x)] / [f(x)]^2
= [1*(x-1) - (x+4)*1] / [(x-1)^2]
= [(x-1 - x - 4)] / [(x-1)^2]
= (-5) / [(x-1)^2]
Therefore, the derivative of g(x) is (-5) / [(x-1)^2].