Differential Calculus

function composition, limits

infinity is a concept, not a number, so imagine a really, really large number

create a graph and/or table

f(x) = 1/x x = 0 f(x) = undefined

lim (x to -3) (-1 |x+3|) / (x+3) no solution at -3

lim (x to inf) (x^2+3) / x^3 = 0 x^3 grows faster than x^2

lim (x to 0) sin(x)/x = 1

dx is h in many literatures

change in something (distance) / with respect to something else (time)

Tangent lines are important because they only touch the function at one point. This point is also a slope of that point and the goal is to find slopes of functions at any point.

Formal definition of a derivative \(\frac{dy}{dx} = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

Derivative of a constant \(\frac{d}{dx} (c) = 0\)

Derivative of a variable \(\frac{d}{dx} x = 1\) \(\frac{d}{dx} x^n = nx^{n-1}\)

Sum rule

Product rule If both f(x) and g(x) are differentiable, then \(\frac{d}{dx} [f(x)g(x)] = f(x)\frac{d}{dx}[g(x)] + g(x)\frac{d}{dx}[f(x)] = f(x)g'(x) + g(x)f'(x)\)

Quotient rule \((\frac{f(x)}{g(x)})' = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2}\)

Chain rule \(\frac{d}{dx} (u \circ v) = \frac{dv}{dx} (\frac{du}{dx} \circ v)\) or \(\frac{du}{dx} = \frac{du}{dv} \frac{dv}{dx}\)

Implicit differentiation is just a special case of chain rule for derivative. Take derivative, add dy/dx where needed.

Related rates; solved with implicit differentiation

An example is, if a balloon is inflated, the rate of volume is increasing but so is the rate of radius, hence related rates.

It’s usually easier to measure the change in one over the others, so once you have the easier one figured out, you can use the solution for that to solve the harder one.

If the volume (V) is \(\frac{dV}{dt} = 100 cm^3/s\) and the radius r and the goal is to find out the radius’s rate when r = 25 cm.

\(V = \frac{4}{3} \pi r^3\) \(\frac{dV}{dt} = \frac{4}{3} \pi 3r^2 \frac{dr}{dt}\) \(\frac{dr}{dt} = \frac{dV}{dt} \frac{1}{4 \pi r^2}\)

Plug in r and evaluate.

Related rates steps:

  1. read the problem carefully and identify all the quantities
  2. use appropriate variables to represent quantities (area, volume, similar triangles, Pythagorean theorem, etc.)
  3. write the equation that relate only changing quantities where rate of change are known and unknown (one unknowns in single variable calculus classes)
  4. this might be a function of time (d/dt) this might involve implicit differentiation
  5. in the resulting equation, substitute quantities back in
  6. solve for the unknown and include any appropriate units of measurements

Rolle’s Theorem

mean value theorem

L’Hopital’s rule Try out by differentiating the top and bottom of fraction when numerator and denominator evaluates to 0 in a limit

Epsilon delta \((\epsilon, \delta)\) definition of limit

Exponential derivative \(\frac{d}{dx} e^{f(x)} = e^{f(x)} * f'(x)\)

Trigonometric derivative \(\frac{d}{dx} sin(u) = cos(u) \frac{du}{dx}\) \(\frac{d}{dx} cos(u) = -sin(u) \frac{du}{dx}\) \(\frac{d}{dx} tan(u) = sec^2(u) \frac{du}{dx}\)

Higher order derivatives Each derivative tells you a short story.

  • Function tells you displacement.
  • First derivative gives you velocity.
  • Second derivative gives you acceleration.
  • Third derivative gives you jerk.

exponential decay: M’(t) = -rM(t) “Removal rate is proportional to the amount of medicine in the bloomstream.” = “rate of change proportional to amount”

Integrate to get d/dx M(t) = -rM(t)

M(0)=d

exponential growth y = Ce^(kt)

Calculus 1 Problems and Solutions