Corrections and explanations week 2

1 files changed, 11 insertions, 11 deletions

diff --git a/assignment2.tex b/assignment2.tex
index fa70c40..f5864bc 100644
--- a/assignment2.tex
+++ b/assignment2.tex
@@ -84,25 +84,25 @@
         \end{enumerate}
 
     \item \begin{enumerate}
-            \item $f'(x) = 4x^3-6x^2.$
-            \item $f'(x) = \frac{(x-7)(2x)-(x^2+5)(1)}{(x-7)^2} = \frac{x+2}{x-7}.$
-            \item \begin{align*}
+            \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 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)' \\
                           &= 2\cdot\sin\sqrt x\cdot\cos\sqrt x\cdot\frac12\cdot x^{-\frac12} \\
                           &= \frac12\sin\left(2\sqrt x\right)\cdot\frac1{\sqrt x} \\
                           &= \frac{\sin\left(2\sqrt x\right)}{2\sqrt x}.
                 \end{align*}
-            \item %todo
-            \item $f'(x) = \left(e^{\tan x}\right)' = e(\tan x)\cdot\tan'x = \frac{e(\tan x)}{\cos^2x}$.
-            \item $f'(x) = -\frac1{\cos x}\cdot-\sin x=\frac{\sin x}{\tan x}$.
-            \item \begin{align*}
+            \item We have $f(x) = 1-\cos^2\left(\sqrt x\right) = \sin^2\left(\sqrt x\right)$, so this is the same as 4c.
+            \item Using the chain rule, $f'(x) = \left(e^{\tan x}\right)' = e^{\tan x}\cdot\tan'x = \frac{e^{\tan x}}{\cos^2x}$.
+            \item Using the chain rule, $f'(x) = -\frac1{\cos x}\cdot-\sin x=\tan x$.
+            \item Using the chain rule, \begin{align*}
                     f'(x) &= \frac1{\sqrt{1-(1-2x)^2}}\cdot-2 \\
-                          &= -2\cdot\frac1{\sqrt{4x^2+4x}} \\
-                          &= -2\cdot\frac1{2\sqrt{x^2+x}} \\
-                          &= -\left(x^2+x\right)^{-\frac12}.
+                          &= -\frac2{\sqrt{-4x^2+4x}} \\
+                          &= -\frac2{2\sqrt{-x^2+x}} \\
+                          &= -\left(-x^2+x\right)^{-\frac12}.
                 \end{align*}
-            \item %todo
+            \item Using the chain rule, $f'(x) = 10^{x^2}\cdot\ln10\cdot2x$.
         \end{enumerate}
 
     \item % todo