# How do you calculate the derivative of [int(sqrt(1+(dy/dx)^2)/60 dx)]?

Aug 13, 2015

Notice how the integral is the opposite of a derivative. The first part of the Fundamental Theorem of Calculus states that, and the term "antiderivative" essentially describes the basic function of an integral. Therefore, the derivative of an integral is the integrand. In other words:

$\frac{d}{\mathrm{dx}} \left\{\int \textcolor{g r e e n}{\left[f \left(x\right) + C\right]} \mathrm{dx}\right\} = \frac{d}{\mathrm{dx}} \left[\textcolor{g r e e n}{g \left(x\right)}\right] = \frac{\mathrm{dg}}{\mathrm{dx}} = \textcolor{b l u e}{f \left(x\right) + C}$

where $g \left(x\right) = \int f \left(x\right) \mathrm{dx} + C x$ (and thus $\frac{\mathrm{dg}}{\mathrm{dx}} = f \left(x\right) + C$). $C$ may be zero as well as nonzero.

$\int \left\{\frac{d}{\mathrm{dx}} \left[f \left(x\right) + C\right]\right\} \mathrm{dx} = \int \left[\frac{\mathrm{df}}{\mathrm{dx}} + 0\right] \mathrm{dx} = \textcolor{b l u e}{f \left(x\right) + C}$

(yes, the indefinite integral of zero is a constant)

For this, we would be using the upper case, differentiating an integral.

$\frac{1}{60} \int \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

which, by the way, is the equation for 1/60th times an arc length over an interval $\left[a , b\right]$.

If we were to try to evaluate this integral manually, even with something like $f \left(x\right) = {x}^{2}$, it would be very difficult. Thankfully, all we need to write out is:

$\frac{d}{\mathrm{dx}} \left[\frac{1}{60} \int \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}\right]$

$= \frac{1}{60} \cdot \frac{d}{\mathrm{dx}} \left[\int \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}\right]$

$= \textcolor{b l u e}{\frac{1}{60} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}}}$

Here, $C = 0$, so that's what we get.