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Differentiation

Differentiation

### Derivative No. 4

Using the methods in this post, I would like to evaluate $\frac{dz}{d\phi}\bigg|_{\phi^+}$ with $z=r\cos\theta$. Substituting, the expression to evaluate is $\frac{d (r \cos\theta )}{d\phi} \bigg|_{\phi^+}$. Since $r \cos\theta$ does not depend on $\phi$, $r \cos\theta$ is a constant function with respect to $\phi$. From this post, it follows that $\boxed { \frac{dz}{d\phi} \bigg|_{\phi^+} = 0 }$.

Differentiation

Differentiation

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 Each $x$ in the product rule has been replaced by $r$.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

### The Crux of Calculus

Define $\Delta x \equiv x_2 – x_1$, to be consistent with this post. Similarly, define $\Delta y \equiv y_2 – y_1$ and $\Delta z \equiv z_2 – z_1$. The Cartesian coordinates are $x$, $y$, & $z$. In contrast, the spherical coordinates are $r$, $\theta$, & $\phi$. Here, $\phi$ is the azimuthal angle in the $xy$-plane. Next, use this post to obtain the equations relating Cartesian coordinates to spherical coordinates. In particular: $x = r \sin\theta \cos \phi$ $y = r \sin \theta \sin \phi$ $z = r \cos\theta$ Note that $x$ changes if $r$ changes, $\theta$ changes, and/or $\phi$ changes. …

Differentiation

### Definition of a Derivative

I copy the definitions of three different types of derivatives from [1]: $\lim_{\Delta x \rightarrow 0} \frac{f(a+\Delta x) – f(a)}{\Delta x} \equiv \frac{d f(x)}{dx}\big|_{a}$ $\lim_{\Delta x \rightarrow 0^+} \frac{f(a+\Delta x) – f(a)}{\Delta x} \equiv \frac{d f(x)}{dx}\big|_{a^+}$ $\lim_{\Delta x \rightarrow 0^-} \frac{f(a+\Delta x) – f(a)}{\Delta x} \equiv \frac{d f(x)}{dx}\big|_{a^-}$ These definitions are best understood geometrically in terms of secant and tangent lines. Note that $\Delta x$ never equals zero because otherwise the fraction would be undefined. The tangent line intersects the function at only one point. [1] David V. Widder. Advanced Calculus. Dover 1989.