Derivatives of trigonometric functions

Correction about \(G(t)=\frac{1-2t}{3+t}\).

  • I should have really calculated those infinite limits
    • Let's look at the situation as \(t\to(-3)^-\).
    • The nominator \(1-2t\) has a nonzero, positive value at \(t=-3\).
    • As \(t\) approaches \((-3)\) from the left, the denominator \(3+t\) gets negative values indefinitely close to zero.
    • Therefore, the fractional \(G(t)=\frac{1-2t}{3+t}\) gets indefinitely small as \(t\to(-3)^-\), which means \(\lim_{t\to(-3)^-}G(t)=-\infty\).
    • The shape of the graph confirms this.

Derivatives of \(\sin(x)\) and \(\cos(x)\)

  • If you're interested, you can find a geometric argument for these derivatives in the textbook.
    • We'll apply Euler formula magic, which will be much quicker!
    • Of course, you just have to remember the formulas themselves, but you have to see how much shorter this proof is.
    • Recall that we have the imaginary number \(i\), which has the wonderful property \(i^2=-1\).
    • Euler's formula was the following. For a real number \(x\), we have that \[ e^{ix}=\cos(x)+i\sin(x). \]
    • Let's derivate this exponential function.
    • In general, for a constant \(a\), we have \(\frac{\mathrm d}{\mathrm dx}e^{ax}=ae^{ax}\).
    • In this particular case, we get \[ (e^{ix})'=ie^{ix}. \]
    • Applying Euler's formula on the left side, we get \[ (e^{ix})'=(\cos(x)+i\sin(x))'=\cos'(x)+i\sin'(x). \]
    • On the right side, we get \[ ie^{ix}=i(\cos(x)+i\sin(x))=i\cos(x)+i^2\sin(x)=i\cos(x)-\sin(x). \]

Derivatives of \(\sin(x)\) and \(\cos(x)\)

  • We equate what we've gotten on the two sides to get: \[ \cos'(x)+i\sin'(x)=-\sin(x)+i\cos(x). \]
    • VoilĂ ! Equating the real parts gives \[ \cos'(x)=-\sin(x), \] and equating the imaginary parts gives \[ \sin'(x)=\cos(x). \]
  • Note that these derivatives give limits we couldn't verify before.
    • Recall that we've made the guess \(\lim_{x\to0}\frac{\sin(x)}{x}=1\) based on numerical evidence.
    • Note that we have \[ \lim_{x\to 0}\frac{\sin(x)}{x}=\sin'(x=0). \]
    • And now we know that. \[ \sin'(x=0)=\cos(x=0)=1. \]
    • Exercise. Find \(\lim_{x\to0}\frac{\cos(x)-1}{x}\).

Derivatives of the other trigonometric functions.

  • Recall the quotient rule: \[ \left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}. \]
    • Using this, we get the following. \[ \tan'(x)=\left(\frac{\sin(x)}{\cos(x)}\right)'=\frac{\sin'(x)\cos(x)-\sin(x)\cos'(x)}{\cos(x)^2} =\frac{\cos(x)^2+\sin(x)^2}{\cos(x)^2}=\frac{1}{\cos(x)^2}. \]
    • Exercises. 3.3. 2, 4, 12, 16, 34, 35, 40, 42, 44, 50