Cheatsheet

Product Rule

$\frac{d}{dx}(f(x)g(x))=f^{\prime }(x)g(x)+g^{\prime }(x)f(x)$

Quotient Rule

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f^{\prime }(x)g(x)-g^{\prime }(x)f(x)}{(g(x){)}^2}$

Chain Rule

Let $f(u)$ be differentiable at $u=g(x)$, and let $g$ be a differentiable function of $x$ $\frac{d}{dx}(f(g(x)))=f^{\prime }(g(x))\cdot g^{\prime }(x)$

Continuous Function

A Function is continuous at a point $c$ if: ${\lim }_{x\to c}f(x)=f(c)$ Note that ${\lim }_{x\to c}f(x)$ has to exist (Laws of Limits)

Implicit Differentiation

1. Differentiate both sides of the equation W.R.T. $x$
   • $\frac{d}{dx}g(y)=g^{\prime }(y)\frac{dy}{dx}$
2. Solve the resulting equation for $\frac{dy}{dx}$

Example

Find $\frac{dy}{dx}$ for $x^3+y^3-9xy=0$ Differentiating both sides W.R.T. $x$, $3x^2+3y^2\frac{dy}{dx}-9y-9x\frac{dy}{dx}=0$ Solving for $\frac{dy}{dx}$, $(3y^2-9x)\frac{dy}{dx}=-3x^2+9y$ $\Rightarrow \frac{dy}{dx}=-\frac{3x^2-9y}{3y^2-9x}$

Laws of Limits

These are true provided all the limits involved exist

\lim_{x \to c} (f(x) \pm g(x)) = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$$ $$\lim_{x \to c} kf(x) = k \lim_{x \to c} f(x)$$ $$\lim_{x \to c} (f(x)g(x)) = \left(\lim_{x \to c}f(x)\right)\left(\lim_{x \to c}g(x)\right)$$ $$\lim_{x \to c}\frac{f(x)}{g(x)} = \frac{\lim\limits_{x \to c} f(x)}{\lim\limits_{x \to c} g(x)}$$ Provided that $\lim\limits_{x \to c}g(x) \neq 0$. If this is an issue, [[Replacement Rule]] can be used If $g(x)$ is continuous at $x = b$, and $\lim\limits_{x \to c} f(x) = b$, then $$\lim_{x \to c}g(f(x)) = g\left(\lim_{x \to c} f(x)\right)$$ $\lim_{x \to c} f(x)$ only exists if: $$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$$ ### Limits of Polynomial Fractions For limits in the form $$\lim_{x \to \pm \infty} \frac{P(x)}{Q(x)}$$ where $P(x)$ and $Q(x)$ are polynomials in $x$. $$\lim_{x \to \pm \infty} \frac{P(x)}{Q(x)}$$ $$= \lim_{x \to \pm \infty}\frac{Ax^\alpha + ...}{Bx^\beta + ...} = \begin{cases} 0 & \text{if }\alpha < \beta \\ \frac{A}{B} & \text{if }\alpha = \beta\\ \infty \ \text{or} \ -\infty & \text{if }\alpha > \beta \end{cases}$$ If $\alpha > \beta$, the sign of the limit depends on the sign of $A$ and $B$ $$\lim_{x \to \infty} \frac{(18x^2 + 5x - 1)(2\sqrt{x}-1)^3}{(3x-1)^4}$$ $$=\lim_{x \to \infty} \frac{144x^\frac{7}{2} + ...}{81x^4 + ...} = 0$$ ### Replacement Rule Let $I$ be an open interval containing the point $x = c$ If $f(x) = g(x)$ for all $I$, except possible at $x = c$, then $$\lim_{x \to c}f(x) = \lim_{x \to c}g(x)$$ #### Eg 1 $$\lim_{x \to 6} \frac{x^2 -7x+6}{36-x^2} = \lim_{x \to 6} \frac{(x-1)(x-6)}{-(6+x)(x-6)} = \lim_{x \to 6} \frac{(x-1)}{-(6+x)} = -\frac{5}{12}$$ This is because $\frac{x-1}{-(6+x)} = \frac{(x-1)(x-6)}{-(6+x)(x-6)}$ except at $x=6$, so we can use $\frac{x-1}{-(6+x)}$ to evaluate $\lim\limits_{x \to 6}$ #### Eg 2 $$\lim_{x \to -3} \frac{\sqrt{x+12} - \sqrt{6-x}}{9-x^2}$$ $$=\lim_{x \to -3} \frac{\sqrt{x+12} - \sqrt{6-x}}{(3+x)(3-x)}\ \cdot\ \frac{\sqrt{x+12} + \sqrt{6-x}}{\sqrt{x+12} + \sqrt{6-x}}$$ $$=\lim_{x \to -3} \frac{(x+12) - (6-x)}{(3+x)(3-x)\left(\sqrt{x+12} + \sqrt{6-x}\right)}=\lim_{x \to -3} \frac{6+2x}{(3+x)(3-x)\left(\sqrt{x+12} + \sqrt{6-x}\right)}=$$ $$=\lim_{x \to -3} \frac{2}{(3-x)\left(\sqrt{x+12} + \sqrt{6-x}\right)} =\frac{2}{(3+3)\left(\sqrt{12-3} + \sqrt{6+3}\right)} =\frac{1}{18}$$