Day 11, Mat129: Calculus I

• Rules that help us avoid having to use the definition each time (with Proofs):
  • The constant multiple rule (makes sense)
    $(cf(x))'=cf'(x)$

  • The sum rule

    works as we'd hope: "The derivative of a sum is the sum of the derivatives."

    $(f(x) + g(x))'=f'(x)+g'(x)$

  • The product rule (doesn't work the way we'd hope):
    $(f(x) \cdot g(x))'=f'(x)g(x)+f(x)g'(x)$

    To prove it we use one of my favorite tricks: the addition of a special form of 0.

  • The power rule:
    $(x^n)'=n x^{n-1}$

    where $n$ is a positive integer.

    We can prove this via dominoes -- i.e., mathematical induction -- or via the binomial theorem)

  • NOTE: at this point we have all the rules necessary to differentiate all polynomials, without needing to resort to the definition!

    The derivative of the monomial $x^n$ is $nx^{n-1}$, and

    The derivative of the monomial $c x^n$ is $nc x^{n-1}$ (by constant multiple).

    A polynomial is just a sum of these. So we apply the sum rule, and the power rule, and the constant multiple rule to the flight of the eraser, to get

    $s'(t)=(at^2+bt+c)'=2at+b$

    and

    $s''(t)=(2at+b)'=2a$

    The general rule for a polynomial $P(x)$, where is \[ P(x)=\sum_{k=0}^n c_{k}x^k \] is \[ P'(x)=\left(\sum_{k=0}^n c_{k}x^k\right)'=\sum_{k=0}^n (c_{k}x^k)'=\sum_{k=1}^n c_{k}kx^{k-1} \] Notice that the 0-term disappears, because it's a constant, and a constant's derivative is 0 (so that term "disappears").

• Other rules:
  • The quotient rule:
    low d high minus high d low,
    over the denominator squared they go.

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

  • The power rule actually works even for non-integers:
    $(x^n)'=n x^{n-1}$

    where $n$ is any non-zero real number (although we won't prove that).

• #13, p. 136
• #22, p. 136 (expand first)
• #48, p. 137
• #58, p. 137

• Radian measure: there is a linear relationship between radians and degrees: $Radians=\frac{2\pi}{360}Degrees$. In calculus, we almost always use radians. You should try to start thinking in terms of radians, too.

• Length of a sector, related to angle $\theta$ subtending and radius $r$:

  Arc Length: $L = \theta r$

  Again, this is an example of a linear relationship: $L$ is proportional to each of $\theta$ and $r$. So

  • if you double the radius, you double the arc length;
  • if you double the angle, you double the arc length.
  

  We can use this diagram (and the arc length formula) to prove a really important limit: \[ \lim\limits_{\theta \to 0}{\frac{\sin(\theta)}{\theta}}=1 \]

  (See page 142, which I'll walk you through now. The proof uses some elementary trigonometry, plus the squeeze theorem, which is one of the really cool things about this proof!)

• Definitions of the trig functions, as functions of $x$, in radians (which is how we think about them in calculus):

We will derive the derivative of the sine function from the limit definition of the derivative (although we can see the derivative graphically above).

There are exactly three important trig identities one needs to know (all the others can be derived from these three):

• The Pythagorean theorem:
  $\sin^2(x)+\cos^2(x)=1$

• The sine of a sum:
  $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$

• The cosine of a sum:
  $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

So let's see how to derive the derivative of the sine from the MIDIC -- the Most Important Definition In Calculus -- the limit definition of the derivative -- using the second of these identities, plus one more important limit: \[ \lim\limits_{\theta \to 0}{\frac{\cos(\theta)-1}{\theta}}=0 \]

Then the derivative of the cosine can be derived by simply shifting the sine function, and using the second trig identity above.

• Derivative of tangent (use the quotient rule, of course!)
• Fourth derivatives of sine, cosine
• #7, p. 146
• #26
• #35
• #49