Calculate the value of the differential of each function

Calculate the value of the differential of each function over the indicated change of the variable. 1° $y = \dfrac{1}{(\tan x+1)^{2}}$, from $x = \dfrac{\pi}{6}$ to $x = \dfrac{61\pi}{360}$; 2° $y = \sin 2\varphi$, from $\varphi = \dfrac{\pi}{6}$ to $\varphi = \dfrac{61\pi}{360}$; 3° $y = \cos^{2}\varphi$, from $\varphi = 60^{\circ}$ to $\varphi = 60^{\circ}30'$.

Step 1: The common increment

All three parts use the same step. For 1° and 2°: \[\Delta x = \frac{61\pi}{360}-\frac{\pi}{6} = \frac{61\pi-60\pi}{360} = \frac{\pi}{360} \approx 0.0087266.\]

For 3° the step is $30' = \tfrac12{}^{\circ}$, which in radians is

\[ \Delta\varphi = \frac12\cdot\frac{\pi}{180} = \frac{\pi}{360}, \]

the same number. Converting to radians is essential: the derivative formulas for trigonometric functions are valid only in radian measure.

Step 2: Part 1°: the derivative

Write $y = (\tan x+1)^{-2}$ and apply the chain rule, with $(\tan x)' = \sec^{2}x$: $y' = -2\bigl(\tan x+1\bigr)^{-3}\sec^{2}x$.

Step 3: Part 1°: evaluate at $x_{0} = \dfrac{\pi}{6}$

At this point $\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt3}$ and $\sec^{2}\dfrac{\pi}{6} = \dfrac{4}{3}$. For the cube in the denominator: \[\frac{1}{\sqrt3}+1 = \frac{1+\sqrt3}{\sqrt3}, \qquad \left(\frac{1+\sqrt3}{\sqrt3}\right)^{3} = \frac{\bigl(1+\sqrt3\bigr)^{3}}{3\sqrt3},\]

\[ \bigl(1+\sqrt3\bigr)^{3} = 1+3\sqrt3+9+3\sqrt3 = 10+6\sqrt3. \]

Therefore \[y'\Bigl|_{\pi/6} = -2\cdot\frac{3\sqrt3}{10+6\sqrt3}\cdot\frac43 = -\frac{8\sqrt3}{10+6\sqrt3} = -\frac{4\sqrt3}{5+3\sqrt3}.\]

Step 4: Part 1°: rationalise and finish

Multiplying above and below by $5-3\sqrt3$, the denominator becomes $25-27 = -2$: \[y'\Bigl|_{\pi/6} = -\frac{4\sqrt3\bigl(5-3\sqrt3\bigr)}{-2} = 2\sqrt3\bigl(5-3\sqrt3\bigr) = 10\sqrt3-18 \approx -0.6794919.\]

\[ dy_{1} = \bigl(10\sqrt3-18\bigr)\cdot\frac{\pi}{360} \approx -0.0059297. \]

\[dy_{1} = \dfrac{\bigl(10\sqrt3-18\bigr)\pi}{360} \approx -0.00593\]

Step 5: Part 2°

Here $y' = 2\cos 2\varphi$, and at $\varphi_{0} = \dfrac{\pi}{6}$ the argument is $2\varphi_{0} = \dfrac{\pi}{3}$, where $\cos = \dfrac12$: \[y'\Bigl|_{\pi/6} = 2\cdot\frac12 = 1, \qquad dy_{2} = 1\cdot\frac{\pi}{360}.\]

\[dy_{2} = \dfrac{\pi}{360} \approx 0.00873\]

Step 6: Part 3°

Using $\bigl(\cos^{2}\varphi\bigr)' = -2\cos\varphi\sin\varphi = -\sin 2\varphi$, at $\varphi_{0} = 60^{\circ}$: \[y'\Bigl|_{60^{\circ}} = -\sin 120^{\circ} = -\frac{\sqrt3}{2}, \qquad dy_{3} = -\frac{\sqrt3}{2}\cdot\frac{\pi}{360} = -\frac{\sqrt3\,\pi}{720}.\]

\[dy_{3} = -\dfrac{\sqrt3\,\pi}{720} \approx -0.00756\]

Step 7: Verification

\[1^{\circ}:\ y\left(\frac{\pi}{6}\right) \approx 0.4019238,\ y\left(\frac{61\pi}{360}\right) \approx 0.3960296; \quad \Delta y \approx -0.0058942 \text{ vs } dy_{1} \approx -0.0059297 \;\;\checkmark\]

\[2^{\circ}:\ \sin\frac{61\pi}{180}-\sin\frac{\pi}{3} \approx 0.0085943 \text{ vs } dy_{2} \approx 0.0087266 \;\;\checkmark\]

\[3^{\circ}:\ \cos^{2}60.5^{\circ}-\cos^{2}60^{\circ} \approx -0.0075190 \text{ vs } dy_{3} \approx -0.0075575 \;\;\checkmark\]