From 93440abf3ab4b3589741c02c586ab89c4c76c831 Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Thu, 10 Sep 2015 21:02:42 +0200
Subject: Correction & finish week 2

---
 assignment2.tex | 36 ++++++++++++++++++++++++++++++++++--
 1 file changed, 34 insertions(+), 2 deletions(-)

(limited to 'assignment2.tex')

diff --git a/assignment2.tex b/assignment2.tex
index f5864bc..44b8c33 100644
--- a/assignment2.tex
+++ b/assignment2.tex
@@ -85,7 +85,7 @@
 
     \item \begin{enumerate}
             \item Using basic rules, $f'(x) = 4x^3-6x^2$.
-            \item Using the division rule, $f'(x) = \frac{(x-7)(2x)-(x^2+5)(1)}{(x-7)^2} = \frac{x+2}{x-7}$.
+            \item Using the division rule, $f'(x) = \frac{(x-7)(2x)-(x^2+5)(1)}{(x-7)^2} = \frac{x^2-14x-5}{x-7}$.
             \item Using the product rule, \begin{align*}
                     f'(x) &= \left(\sin{\sqrt x}\cdot\sin{\sqrt x}\right) \\
                           &= 2\cdot\sin\sqrt x\cdot\left(\sin\sqrt x\right)' \\
@@ -105,7 +105,39 @@
             \item Using the chain rule, $f'(x) = 10^{x^2}\cdot\ln10\cdot2x$.
         \end{enumerate}
 
-    \item % todo
+    \item \begin{enumerate}
+            \item Using the chain rule, $f'(x) = e^{\sin x}\cdot\cos x$.
+            \item Using logarithmic differentiation:
+
+                \begin{align*}
+                    f(x)        &= \left(e^x\right)^{e^x} \\
+                    \ln(f(x))   &= \ln\left(\left(e^x\right)^{e^x}\right) = e^x\cdot x \\
+                    \frac{f'(x)}{f(x)} &= e^x + x\cdot e^x = (x+1)\cdot e^x \qquad\text{(product rule)}\\
+                    f'(x)       &= f(x) \cdot (x+1)\cdot e^x \\
+                                &= \left(e^x\right)^{e^x+1}\cdot(x+1).
+                \end{align*}
+            \item First we rewrite in order to get the inverse:
+
+                \begin{align*}
+                    y       &= e^{2x} \\
+                    \ln y   &= 2x \\
+                    x       &= \frac{\ln y}2 \\
+                    f^{-1}(x) &= \frac12\ln x.
+                \end{align*}
+
+                Next, we find the derivative as $\left(f^{-1}\right)'(x) = \frac12\cdot\frac1x=\frac1{2x}$.
+
+            \item Again, we first find the inverse:
+
+                \begin{align*}
+                    y           &= \sqrt{x-2} \\
+                    y^2         &= x-2 \\
+                    y^2 + 2     &= x \\
+                    f^{-1}(x)   &= x^2 + 2.
+                \end{align*}
+
+                Then it's easy to see $\left(f^{-1}\right)'(x) = 2x$.
+        \end{enumerate}
 \end{enumerate}
 
 \end{document}
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