Derivative No. 1

 I would like to evaluate

$ \frac{dx(r)}{dr} \bigg|_{x^+}$


$x(r) = r \sin\theta \cos\phi$.

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Each $x$ in the product rule has been replaced by $r$.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

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


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

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

Christina Daniel

I am a research assistant in theoretical physics. This website,, serves to organize my ongoing learning and research as well as to provide a resource to other learners.

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