Problem Set 08s

# F x lnln ln x solution cos x f x lnln ln

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Unformatted text preview: ot x The domain of the function is valid for y x 2 + ln x &gt; 0 ' ln(sin x ) * ln x &gt; −2 y ' = y) + ( ln x ) cot x , ( + x Domain: x &gt; e −2 ln x ⎡ྎ ln(sin x ) + x ln x cot x ⎤ྏ y ' = (sin x ) ⎢ྎ ⎥ྏ x b. f ( x) = ln[ln (ln x )] ⎣ྏ ⎦ྏ Solution: cos x f ( x) = ln[ln (ln x )] y = (ln x ) Solution: y = ( ln x ) 1 ⎛ྎ 1 ⎞ྏ⎛ྎ 1 ⎞ྏ ⎜ྎ ⎟ྏ⎜ྎ ⎟ྏ ln (ln x ) ⎝ྎ ln x ⎠ྏ⎝ྎ x ⎠ྏ 1 f ' ( x) = x(ln x ) ln (ln x ) f ' ( x) = cos x cos x ln y = ln !( ln x ) # = cos x ln ( ln x ) &quot; \$ &amp; 1 )&amp; 1 ) 1 y ' = − sin x ln ( ln x ) + cos x ( +( + ' ln x *' x * y cos x = − sin x ln ( ln x ) x ln x ⎤ྏ cos x ⎡ྎ cos x y ' = (ln x ) ⎢ྎ − sin x ln (ln x )⎥ྏ x ln x ⎣ྏ ⎦ྏ The domain of the function: ln (Z ) ⇒ Z &gt; 0 ln (ln x ) &gt; 0 ⇒ ln (x ) &gt; 1 x&gt;e 4. Find the equation of the tangent line to function at prescribed point a. ln ( xy ) − 2 x = 0 d. x y = y x Solution: xy = yx y ln x = x ln y !1\$ !1\$ y ' ln x + y # &amp; = ln y + x # &amp; y ' &quot;x% &quot; y% ! ! y ln x − x \$ x\$ y y ' # ln x − &amp; = ln y − ⇒ y ' # &amp; y% x y &quot; &quot; % x ln y − y = x xy ln y − y 2 y' = xy ln x − x 2 3. Differentiate f and find the domain of f 2 Hung Nguyen b. Problem Set 08 Solution 3.6 Derivatives of Logarithmic Functions y = x 2 ln x, ( x, y ) = (1, 0) Solution: y = x 2 ln x !1\$ y ' = 2 x ln x + x 2 # &amp; = 2 x ln x + x &quot;x% = 2(1) ln(0 ) + 1 = 1 y = x − 1 x =1 3 GE 207C...
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## This document was uploaded on 02/27/2014 for the course M 408c at University of Texas at Austin.

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