Differences

This shows you the differences between two versions of the page.

--- book:chap3:3.2_time_derivatives_of_vectors [2021/11/18 01:59] – [3.2.1 Self Test] jv
+++ book:chap3:3.2_time_derivatives_of_vectors [2021/11/18 02:03] (current) – [3.2.1 Self Test] jv
@@ Line 85: / Line 85: @@
     \frac{\mathrm{d}}{\mathrm{d} x} \ln  x &= x^{-1}
 \end{align*}
-Use only the three rules for derivatives
+and use only the three rules for derivatives
 \begin{align*}
     \frac{\mathrm{d}}{\mathrm{d} x} \bigl( f(x) + g(x) \bigr)
@@ Line 98: / Line 98: @@
 to work out the following derivatives
-**a)** $\displaystyle \sinh x = \frac{1}{2} \left( \mathrm{e}^x - \mathrm{e}^{-x} \right) \quad$ and $\quad \displaystyle \cosh x = \frac{1}{2} \left( \mathrm{e}^x + \mathrm{e}^{-x} \right)$\\
+<WRAP group>
-**b)** $\cos x = \sin(\pi/2 + x) \qquad$\\
+<WRAP half column>
-**c)** $x^a = \mathrm{e}^{a \, \ln x}$ for $a \in \mathbb{R}$. What does this imply for the derivative of $f(x) = x^{-1}$?\\
+**a)** $\displaystyle \sinh x = \frac{1}{2} \left( \mathrm{e}^x - \mathrm{e}^{-x} \right) \quad$ and $\quad \displaystyle \cosh x = \frac{1}{2} \left( \mathrm{e}^x + \mathrm{e}^{-x} \right)$
+**b)** $\cos x = \sin(\pi/2 + x) \qquad$
+**c)** $x^a = \mathrm{e}^{a \, \ln x}$ for $a \in \mathbb{R}$. \\
+$\quad$ What does this imply for the derivative of $f(x) = x^{-1}$?
+</WRAP>
+<WRAP half column>
 **d)** Use the result from c) to proof the quotient rule:
 \begin{align*}
     \frac{\mathrm{d}}{\mathrm{d} x} \frac{ f(x) }{ g(x) }
     = \frac{ f'(x) \, g(x) - f(x) \, g'(x) }{ \bigl( g(x) \bigr)^2 }
-\end{align*}\\
+\end{align*} \\
-**e)** $\displaystyle \tan x = \frac{ \sin x }{ \cos x } \quad $ and $\quad \displaystyle \tanh x = \frac{ \sinh x }{ \cosh x }$\\
+**e)** $\displaystyle \tan x = \frac{ \sin x }{ \cos x } \quad $ and $\quad \displaystyle \tanh x = \frac{ \sinh x }{ \cosh x }$ \\
 **f)** :!: Find the derivative of $\ln x$ solely based on
 $\displaystyle \frac{\mathrm{d}}{\mathrm{d} x} \mathrm{e}^x = \mathrm{e}^x$.
 \\
-++Hint: | Use that $x = \mathrm{e}^{\ln x}$ and take the derivative of both sides.++
+$\quad$ ++Hint: | Use that $x = \mathrm{e}^{\ln x}$ and take the derivative of both sides.++
+</WRAP>
+</WRAP>