derivative

Derivative No. 3

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= r \frac{d( \cos\theta )}{dr}\big|_{r^+} + \cos\theta \frac{dr}{dr}\big|_{r^+}$ Each $x$ in the product rule has been replaced by $r$.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 …

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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 …

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 Each $x$ in the product rule has been replaced by $r$.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

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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, …

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