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Differentiation

How to Use the Product Rule

I would like to evaluate two more derivatives. They are $ \frac{dx}{d\phi}\big|_{\phi^+} $ and $ \frac{dz}{d\theta}\big|_{\theta^+} $ given $x = r \sin\theta\cos\phi$ and $z = r\cos\theta$.   Start with $ \frac{dx}{d\phi}\big|_{\phi^+} $. The first step is to substitute $x$ with $r \sin\theta \cos\phi$. $ \frac{dx}{d\phi}\big|_{\phi^+} = \frac{d r \sin\theta \cos \phi}{d\phi}\big|_{\phi^+} $. The next step is to apply the product rule to the right side. $\frac{dx}{d\phi}\big|_{\phi^+} = r \sin\theta \frac{d \cos \phi}{d\phi}\big|_{\phi^+} + \cos \phi \frac{dr \sin\theta}{d\phi}\big|_{\phi^+}$. Using this derivative of cosine, the previous equation becomes $\frac{dx}{d\phi}\big|_{\phi^+} = -r \sin\theta \sin\phi + \cos \phi \frac{d r \sin\theta}{d\phi}\big|_{\phi^+}$. From this post about the derivative of …

Differentiation

The Product Rule

In this post, I derive the so-called product rule that is taught in a Calculus course. The product rule enables one to find the derivative of a function which can be expressed as a product of two functions. That is,  the product rule allows for evaluating $ \frac{d h(x)}{dx} \big|_{a^+}$ with $ h(x) \equiv f(x) g(x) $. To start, use the corresponding definition of a derivative from this post : $ \frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{h(a + \Delta x) – h(a)}{\Delta x}$ Substitute $ h(x) \equiv f(x) g(x) $ : $ \frac{d h(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow …