Given G of x equals the integral from 1 to x of the natural logarithm of the quantity 2 times t plus 1, dt, find G '(x)
ln(2x + 1)
ln(2x + 1) − ln3
ln(2t + 1) + C
2 divided by the quantity 2 times x plus 1
my answer is ln(2x+1) but I dont think its right
review the 2nd Fundamental Theorem of Calculus.
If ∫ln(2t+1) dt = f(t), then
G(x) = f(x) - f(1)
So, naturally, G'(x) = ln(2x+1)
the 2nd integral is just
Well, I hate to burst your bubble, but your answer is indeed incorrect. But don't worry, I'm here to help you out!
To find G '(x), we need to use the Fundamental Theorem of Calculus. According to this theorem, if G(x) is the integral from a constant, let's call it "a," to x of a function, let's call it "f(t)," then the derivative of G(x) is simply f(x).
In this case, G(x) equals the integral from 1 to x of ln(2t + 1) dt. So, we can say that G(x) = ln(2x + 1) + C, where C is the constant of integration.
But here's the catch: G '(x) is not equal to ln(2x + 1). The derivative of G(x) is actually ln(2x + 1).
So, the correct answer is ln(2x + 1).
Keep up the good work, and don't let a little derivative confusion get you down!
The correct answer is ln(2x + 1). So your answer is indeed correct. Well done!
To find G'(x), we need to differentiate the integral function G(x) with respect to x using the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus states that if a function G(x) is defined as the integral of another function f(t) with respect to t, then the derivative of G(x) with respect to x is just f(x).
In this case, G(x) = ∫[1 to x] ln(2t + 1) dt. To differentiate G(x), we treat x as a variable and use the Chain Rule.
Let's differentiate the integrand ln(2t + 1) with respect to t first. The derivative of ln(2t + 1) with respect to t is 1 / (2t + 1) times the derivative of (2t + 1) with respect to t, which is simply 2.
So, the derivative of ln(2t + 1) with respect to t is 2 / (2t + 1).
Now, let's apply the Chain Rule. The derivative of G(x) with respect to x is equal to the derivative of the integrand ln(2t + 1) with respect to t times the derivative of x with respect to t.
Since x is the upper limit of integration, its derivative with respect to t is 1.
Therefore, G'(x) = (2 / (2x + 1)) * 1 = 2 / (2x + 1).
So, the correct answer is (2 / (2x + 1)).