# 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Each $x$ in the product rule has been replaced by $r$.

Since $\sin\theta \cos\phi$ does not depend on $r$, $\sin\theta \cos\phi$ is a constant function with respect to $r$. Differentiating the constant function,

$\frac{d ( \sin\theta \cos\phi)}{dr} \bigg|_{r^+} = 0$, so

$\frac{dx(r)}{dr} \bigg|_{r^+} = \sin\theta \cos\phi \frac{dr}{dr} \bigg|_{r^+}$.

From the Calculus post,

$\frac{dr}{dr} \bigg|_{r^+} = \frac{dr |_{r^+}}{dr |_{r^+}} = \frac{ \lim_{\Delta r \rightarrow 0^+} \Delta r }{ \lim_{\Delta r \rightarrow 0^+} \Delta r }$.

From a limit property,

$\frac{ \lim_{\Delta r \rightarrow 0^+} \Delta r }{ \lim_{\Delta r \rightarrow 0^+} \Delta r } = \lim_{\Delta r \rightarrow 0^+} \frac{\Delta r}{\Delta r} = \lim_{\Delta r \rightarrow 0^+} 1 = 1$.

Hence

$\boxed {\frac{dr}{dr} \bigg|_{r^+} = 1 }$ and

$\boxed{ \frac{dx(r)}{dr} \bigg|_{r^+} = \sin\theta \cos\phi }$.

#### Written by Christina Daniel

I am a research assistant in theoretical physics. This website, derive-it.com, serves to organize my ongoing learning and research as well as to provide a resource to other learners around the world.