Use the Fundamental Theorem of Calculus to find the derivative of
g(x)=1/(1+t^5)dt in the interval 1 to x.
To find the derivative of the function g(x) = 1/(1+t^5)dt in the interval 1 to x using the Fundamental Theorem of Calculus, we can follow these steps:
1. First, let's define a new function F(x) as the integral of g(t) with respect to t from the lower limit (1 in this case) to x:
F(x) = ∫[1,x] g(t) dt
2. The Fundamental Theorem of Calculus states that if F(x) is defined as the integral of a function g(t) over an interval [a, x], and if g(t) is continuous on the interval [a, x], then the derivative of F(x) can be expressed as:
F'(x) = g(x)
This means that to find the derivative of F(x), we simply need to evaluate g(x).
3. In our case, g(x) = 1/(1+t^5). So, to find F'(x), we find g(x) by substituting x into the function g(t):
F'(x) = g(x) = 1/(1+x^5)
Therefore, the derivative of g(x)=1/(1+t^5)dt in the interval 1 to x is equal to 1/(1+x^5).