The integral from a to b of f(x)/g(x) equals ln[(g(b)/g(a)]. Find g'(x), and explain your reasoning.
since ln(g(b)/g(a) = ln(g(b)) - ln(g(a))
g'(x) = 1/x
Integral of 1/x dx = ln(x)
I still do not understand how you found g'(x) to be 1/x. Could you please elaborate a little more on the reasoning behind this?
Good question. My reasoning is bogus. I was thinking ln(b)-ln(a)
We know that ∫dg/g = ln(g)
dg = g' dx
so, dg/dx = g' = f
how's that?
For example
∫(2x+3)/(x^2 + 3x - 7) dx
let g = x^2 + 3x - 7
dg = (2x+3) dx
and we have
∫dg/g = ln(g)
evaluate at a and b to get
ln(g(b)) - ln(g(a)) = ln[g(b)/g(a)]
To find g'(x), we can differentiate both sides of the equation using the fundamental theorem of calculus. The fundamental theorem of calculus states that if F(x) is an antiderivative of f(x) on [a, b], then the integral of f(x) from a to b is equal to F(b) - F(a).
Let's start by differentiating ln[(g(b)/g(a))] with respect to x using the chain rule:
d/dx [ln[(g(b)/g(a))]] = 1/[(g(b)/g(a)))] * d/dx[(g(b)/g(a))]
To simplify further, we can rewrite ln[(g(b)/g(a))] as ln(g(b)) - ln(g(a)). Now, the equation becomes:
1/[(g(b)/g(a)))] * d/dx[(g(b)/g(a))] = d/dx[ln(g(b)) - ln(g(a))]
Now, let's focus on differentiating the right side of the equation:
d/dx[ln(g(b)) - ln(g(a))] = d/dx[ln(g(b))] - d/dx[ln(g(a))]
To differentiate ln(g(b)) and ln(g(a)), we can use the chain rule again:
d/dx[ln(g(b))] = 1/g(b) * g'(b)
d/dx[ln(g(a))] = 1/g(a) * g'(a)
Now, we can substitute these results back into the equation:
1/[(g(b)/g(a)))] * d/dx[(g(b)/g(a))] = 1/g(b) * g'(b) - 1/g(a) * g'(a)
We can simplify the left side by rewriting it as g(a)/g(b):
g(a)/g(b) * d/dx[(g(b)/g(a))] = 1/g(b) * g'(b) - 1/g(a) * g'(a)
Multiplying both sides of the equation by g(b) * g(a), we get:
g(a) * d/dx[(g(b)/g(a))] * g(b) = g(b) * g'(b) - g(a) * g'(a)
Now, we can cancel out the g(b) and g(a) terms, if g(b) and g(a) are nonzero:
d/dx[(g(b)/g(a))] = g'(b) - g'(a)
So, we have found that the derivative of g(b)/g(a) with respect to x is equal to g'(b) - g'(a).