# Constant Product rule The constant product rule says that if I have $f(x)$ and know $f'(x)$ and I have some $g(x)$ that is a constant multiple of $f(x)$. ie $g(x) = mf(x),$ then $g'(x) = mf'(x),$ ...or in Leibniz notation $ \frac{d}{dy}mf(x) = m \frac{d}{dy} f(x). $ This can be understood geometrically if we recall that the [[Translation of an arbitrary function|effect of a constant product]] is to dilate (stretch or squeeze) things in the vertical dimension. So every unit of vertical movement prior to this dilation maps on to $m$ units after dilation. This means that any tangent line will be $m$ times steeper in the new system of coordinates than in the old one. Practically speaking this is convenient because it means you can often pull constant coefficients out of terms before differentiating if it makes things simpler. # Sum and Difference rule This rule says (informally) that I can add derivatives together or split an expression of multiple terms and differentiate each part to find the derivative of the whole expression. More formally, suppose I have two differentiable functions $f$ and $g$, then $ \frac{\mathrm{d}}{\mathop{\mathrm{d}x}}(f(x)\pm g(x)) = f'(x) \pm g'(x) $ ...which is mighty convenient when tackling more complex expressions. See [libretext maths](https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/02%3A_Derivatives/2.03%3A_Basic_Differentiation_Rules) for more on the constant multiple rule and the sum and difference rule. # Product and quotient rules This rule says I have two differentiable functions $f$ and $g$, then $ \begin{align*} \frac{\mathrm{d}}{\mathop{\mathrm{d}x}}(f(x)g(x)) &= f(x)g'(x)+g(x)f'(x)\\ \text{and}\frac{\mathrm{d}}{\mathop{\mathrm{d}x}}\frac{f(x)}{g(x)} &= \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2} \end{align*} $ Note that in the second case the order of the functions matters so make sure you get them the right way around. See [libretext maths](https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/02%3A_Derivatives/2.04%3A_The_Product_and_Quotient_Rules) for more detail on these two. # Chain rule and $u$-substitution The chain rule allows us to unpack composite expressions. It says that if we have some function $h$ of the form $h(x) = f(g(x))$, then $ h'(x) = f'(g(x)).g'(x) $ It is easier to understand by means of an example. Suppose we have to find a derivative such as $ \frac{\mathrm{d}}{\mathop{\mathrm{d}x}} \cos(x^2), $ The way to proceed is to define an intermediate function (called $u$ by convention) which we can then use to apply the chain rule. Let's quickly restate the chain rule in terms of $u$ to make this more explicit. If we can rewrite our function into the form $f(u(x))$, then the derivative we want to find will be $f'(u(x))u'(x)$. OK $ \begin{align*} \text{let}~u&=x^2\\ \text{then}~\mathrm{d}u&=2x\;\mathrm{d}x\\ \text{and}~\frac{\mathrm{d}}{\mathop{\mathrm{d}x}} \cos(x^2) &= \frac{\mathrm{d}}{\mathop{\mathrm{d}u}} \cos(u) \;\mathrm{d}u\\ &=-2x\sin(x^2) \end{align*} $ ## Linear multiples and the chain rule Since $\frac{\mathrm{d}}{\mathop{\mathrm{d}x}}ax=a$, a perhaps obvious consequence of the chain rule that is worth just committing to memory is that if $f(x)$ is some differentiable function, then $\frac{\mathrm{d}}{\mathrm{d}x}f(ax)=af'(ax)$. A few examples 1. $ \frac{\mathrm{d}}{\mathop{\mathrm{d}x}}\sin(5x)=5\cos(5x) $ 2. $ \frac{\mathrm{d}}{\mathop{\mathrm{d}x}}\frac{1}{\mathrm{e}^{4x}}=\frac{\mathrm{d}}{\mathop{\mathrm{d}x}}\mathrm{e}^{-4x}=-4\mathrm{e}^{-4x}=-\frac{4}{\mathrm{e}^{4x}} $ These are so common that it shouldn't be something you have to waste time actually doing the substitution especially when you need derivatives to do things like integrating by parts. Read more on [libretext maths](https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/02%3A_Derivatives/2.05%3A_The_Chain_Rule) about the chain rule.