Integral of the derivative of a function

Suppose $\displaystyle f(x)$ is a differentiable function whose derivative is $\displaystyle \dfrac{1}{1+x^2}$. We know that $\displaystyle f(x) = \arctan{x}+C$. Now, suppose $\displaystyle g(t) = \tan(t^3+4t^2-2t+7)$ and $\displaystyle h(t) = \cot(t^3-7t^2-34t+1)$. Suppose we want to figure out what $\displaystyle f(h(t))-f(g(t))$ is. Taking the derivative, we get $\displaystyle f^\prime(h(t))h^\prime(t)-f^\prime(g(t))g^\prime(t) = 6(t+2)(t-3)$. Integrating, we get $\displaystyle 2t^3-3t^2-36t+C$ which undoubtedly achieves values outside the range of $\displaystyle \arctan(x)$ regardless of the constants for each. Is this because $\displaystyle \tan(x)$ is discontinuous?

Re: Integral of the derivative of a function

Hey SlipEternal.

The discontinuity in addition to the vertical asymptotes both contribute to this phenomena. You can't integrate something when you have these kinds of things involved.

Re: Integral of the derivative of a function

I have a doubt about your question:

You have f(x) = arc tan x + C

Now f(h(t)) = arc tan[ cot ( t^3 - 7t^2 - 34 t + 1 )] = arc tan[ tan { pi/2 - ( t^3 - 7t^2 - 34 t + 1 )}] = pi/2 - ( t^3 - 7t^2 - 34 t + 1 )

Similarly f(g(t)) = t^3 + 4t^2 - 2 t + 1

Now you can proceed to evaluate

f(h(t)) - f(g(t))

Re: Integral of the derivative of a function

Quote:

Originally Posted by

**chiro** Hey SlipEternal.

The discontinuity in addition to the vertical asymptotes both contribute to this phenomena. You can't integrate something when you have these kinds of things involved.

So, if I let $\displaystyle A = \{t\in \mathbb{R}\mid n\in \mathbb{Z}, t^3+4t^2-2t+7 = (2n+1)\dfrac{\pi}{2} \}$, then I define $\displaystyle A^+ = \{t\in A \mid t\ge 0\}, A^- = \{t\in A \mid t < 0\}$, I can define an set of points $\displaystyle (a_n)_{n\in \mathbb{Z}}$ where $\displaystyle a_0 = \max A^-, a_{-n} = \max\left(A^- \setminus \{a_0,\ldots, a_{1-n}\}\right)$, $\displaystyle a_1 = \min A^+, a_n = \min\left(A^+ \setminus \{a_1,\ldots,a_{n-1}\}\right)$, I can integrate $\displaystyle f^\prime(g(t))g^\prime(t)$ over each interval $\displaystyle (a_{n-1},a_n)$, and in each case, I will get $\displaystyle t^3+4t^2-2t+C$. I can use a similar trick for $\displaystyle f^\prime(h(t))h^\prime(t)$. So, the integral works, and produces the correct result. But, my resulting function has a lot of points of discontinuity (the union of the points of discontinuity of $\displaystyle g(t)$ and $\displaystyle h(t)$ individually).

Now, $\displaystyle \sec^2 x$ has an infinite number of points of discontinuity, yet when I integrate that, I get $\displaystyle \tan x + C$ (here, there is a single constant that works for each interval). Yet, $\displaystyle f(x)$ and $\displaystyle f^\prime(x)$ both have no points of discontinuity, so the composition functions should only be discontinuous at values of $\displaystyle t$ that are points of discontinuity for $\displaystyle g(t)$ and/or $\displaystyle h(t)$, and each of those discontinuities for the composition functions "should" be removable discontinuities since $\displaystyle \displaystyle \lim_{x\to -\infty}f^\prime(x) = \lim_{x\to \infty} f^\prime(x) = 0$, which further confuses me as to why I get a piecewise continuous function where each piece may have a different constant of integration.

To better see my confusion, consider the function $\displaystyle j(t) = \tan(t)$. Now, $\displaystyle f^\prime(j(t))j^\prime(t) = 1$, and the integral would be $\displaystyle t+C$. So, it is the co-restriction of the domain that is confusing me. Would the actual formula be $\displaystyle f(j(t)) = t - \left\lfloor \dfrac{t+\frac{\pi}{2}}{\pi} \right\rfloor\pi + C$ where $\displaystyle \lfloor \cdot \rfloor$ is the floor function?