Using the methods in this post, I would like to evaluate $\frac{dz(r)}{dr}\bigg|_{r^+}$ with $z(r)=r\cos\theta$. Substituting, the expression to evaluate is $ \frac{d (r \cos\theta)}{dr} \bigg|_{r^+}$. From the product rule, $ \frac{dz(r)}{dr} \bigg|_{r^+} = r \frac{d( \cos\theta )}{dr}\big|_{r^+} + \cos\theta \frac{dr}{dr}\big|_{r^+}$ Each $x$ in the product rule has been replaced by $r$. Since $ \cos\theta $ does not depend on…

## Derivative No. 2

Using the methods in this post, I would like to evaluate $\frac{dy(r)}{dr}\bigg|_{r^+}$ with $y(r)=r \sin\theta\sin\phi$. This method is not the traditional method taught in a Calculus course because I only want to use properties and rules that I have derived myself on this website. Substituting, the expression to evaluate is $ \frac{d ( r \sin…

## Derivative No. 1

I would like to evaluate $ \frac{dx(r)}{dr} \bigg|_{x^+}$ with $x(r) = r \sin\theta \cos\phi$. Substituting, the expression to evaluate is $ \frac{d(r \sin\theta \cos\phi )}{dr} \bigg|_{x^+}$ From the product rule, $ \frac{dx(r)}{dr} \bigg|_{r^+} = \sin\theta \cos\phi \frac{dr}{dr} \bigg|_{r^+} + r \frac{d ( \sin\theta \cos\phi)}{dr} \bigg|_{r^+}$. Each $x$ in the product rule has been replaced by $r$….

## The Derivative of a Constant Function

From this post, one definition of a derivative is $\lim_{\Delta x\rightarrow0^+}\frac{f(a+\Delta x)-f(a)}{\Delta x}\equiv\frac{d f(x)}{dx}\big|_{a^+}$. In this case, the values of $\Delta x$ are restricted to positive values due to the $+$ in $0^+$ written in the limit. A function that does not vary with respect to an independent variable is called a constant function. On a…