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d 3 y d x 3 = 2 ( d 2 y d x 2 ) ( d y d x ) . \frac{\operatorname{d^3}\!y}{\operatorname{d}\!x^3} = 2 \left( \frac{\operatorname{d^2}\!y}{\operatorname{d}\!x^2} \right) \left( \frac{\operatorname{d}\!y}{\operatorname{d}\!x} \right). d x 3 d 3 y = 2 ( d x 2 d 2 y ) ( d x d y ) .
y = 1 2 x 2 + 1 12 x 4 + ⋯ y = \frac{1}{2} x^2 + \frac{1}{12} x^4 + \dotsb y = 2 1 x 2 + 12 1 x 4 + ⋯
1 16 π 2 + 1 1536 π 4 . \frac{1}{16} \pi^2 + \frac{1}{1536} \pi^4. 16 1 π 2 + 1536 1 π 4 .
0.005219. 0.005219. 0.005219. Full solutions
(a)
y = ln sec x e y = sec x \begin{gather*}
y = \ln \sec x \\
\mathrm{e}^y = \sec x
\end{gather*} y = ln sec x e y = sec x Differentiating wrt x , x, x ,
d y d x e y = sec x tan x d y d x e y = e y tan x d y d x = tan x \begin{gather*}
\frac{\operatorname{d}\!y}{\operatorname{d}\!x} \mathrm{e}^y = \sec x \tan x \\
\frac{\operatorname{d}\!y}{\operatorname{d}\!x} \mathrm{e}^y = \mathrm{e}^y \tan x \\
\frac{\operatorname{d}\!y}{\operatorname{d}\!x} = \tan x \\
\end{gather*} d x d y e y = sec x tan x d x d y e y = e y tan x d x d y = tan x Differentiating wrt x , x, x ,
d 2 y d x 2 = sec 2 x = tan 2 x + 1 = ( d y d x ) 2 + 1 \begin{align*}
\frac{\operatorname{d^2}\!y}{\operatorname{d}\!x^2} &= \sec^2 x \\
&= \tan^2 x + 1 \\
&= \left( \frac{\operatorname{d}\!y}{\operatorname{d}\!x} \right)^2 + 1
\end{align*} d x 2 d 2 y = sec 2 x = tan 2 x + 1 = ( d x d y ) 2 + 1 Differentiating wrt x , x, x ,
d 3 y d x 3 = 2 ( d 2 y d x 2 ) ( d y d x ) ■ \begin{align*}
\frac{\operatorname{d^3}\!y}{\operatorname{d}\!x^3} &= 2 \left( \frac{\operatorname{d^2}\!y}{\operatorname{d}\!x^2} \right) \left( \frac{\operatorname{d}\!y}{\operatorname{d}\!x} \right) \; \blacksquare
\end{align*} d x 3 d 3 y = 2 ( d x 2 d 2 y ) ( d x d y ) ■
(b)
Differentiating wrt x , x, x ,
d 4 y d x 4 = 2 ( d 3 y d x 3 ) ( d y d x ) + 2 ( d 2 y d x 2 ) 2 \begin{align*}
\frac{\operatorname{d^4}\!y}{\operatorname{d}\!x^4} &= 2 \left( \frac{\operatorname{d^3}\!y}{\operatorname{d}\!x^3} \right) \left( \frac{\operatorname{d}\!y}{\operatorname{d}\!x} \right) + 2 \left( \frac{\operatorname{d^2}\!y}{\operatorname{d}\!x^2} \right)^2
\end{align*} d x 4 d 4 y = 2 ( d x 3 d 3 y ) ( d x d y ) + 2 ( d x 2 d 2 y ) 2 When x = 0 , x=0, x = 0 ,
y = ln sec 0 = 0 d y d x = tan 0 = 0 d 2 y d x 2 = sec 2 0 = 1 d 3 y d x 3 = 2 ( 1 ) ( 0 ) = 0 d 4 y d x 4 = 2 ( 0 ) ( 0 ) + 2 ( 1 ) 2 = 2 \begin{align*}
y &= \ln \sec 0 \\
&= 0 \\
\frac{\operatorname{d}\!y}{\operatorname{d}\!x} &= \tan 0 \\
&= 0 \\
\frac{\operatorname{d^2}\!y}{\operatorname{d}\!x^2} &= \sec^2 0 \\
&= 1 \\
\frac{\operatorname{d^3}\!y}{\operatorname{d}\!x^3} &= 2 \left( 1 \right) \left( 0 \right) \\
&= 0 \\
\frac{\operatorname{d^4}\!y}{\operatorname{d}\!x^4} &= 2 \left( 0 \right) \left( 0 \right) + 2 \left( 1 \right)^2 \\
&= 2
\end{align*} y d x d y d x 2 d 2 y d x 3 d 3 y d x 4 d 4 y = ln sec 0 = 0 = tan 0 = 0 = sec 2 0 = 1 = 2 ( 1 ) ( 0 ) = 0 = 2 ( 0 ) ( 0 ) + 2 ( 1 ) 2 = 2 Maclaurin expansion of y : y: y :
y = 0 + 0 x + 1 2 ! x 2 + 0 3 ! x 3 + 2 4 ! x 4 + ⋯ = 1 2 x 2 + 1 12 x 4 + ⋯ ■ \begin{align*}
y &= 0 + 0x + \frac{1}{2!} x^2 + \frac{0}{3!} x^3 + \frac{2}{4!} x^4 + \dotsb \\
&= \frac{1}{2} x^2 + \frac{1}{12} x^4 + \dotsb \; \blacksquare
\end{align*} y = 0 + 0 x + 2 ! 1 x 2 + 3 ! 0 x 3 + 4 ! 2 x 4 + ⋯ = 2 1 x 2 + 12 1 x 4 + ⋯ ■
(c)
Let x = π 4 x = \frac{\pi}{4} x = 4 π
ln sec π 4 ≈ 1 2 ( π 4 ) 2 + 1 12 ( π 4 ) 4 ln 2 ≈ 1 32 π 2 + 1 3072 π 4 1 2 ln 2 ≈ 1 32 π 2 + 1 3072 π 4 ln 2 ≈ 1 16 π 2 + 1 1536 π 4 ■ \begin{align*}
\ln \sec \frac{\pi}{4} &\approx \frac{1}{2} \left( \frac{\pi}{4} \right)^2 + \frac{1}{12} \left( \frac{\pi}{4} \right)^4 \\
\ln \sqrt{2} &\approx \frac{1}{32} \pi^2 + \frac{1}{3072} \pi^4 \\
\frac{1}{2} \ln 2 &\approx \frac{1}{32} \pi^2 + \frac{1}{3072} \pi^4 \\
\ln 2 &\approx \frac{1}{16} \pi^2 + \frac{1}{1536} \pi^4 \; \blacksquare
\end{align*} ln sec 4 π ln 2 2 1 ln 2 ln 2 ≈ 2 1 ( 4 π ) 2 + 12 1 ( 4 π ) 4 ≈ 32 1 π 2 + 3072 1 π 4 ≈ 32 1 π 2 + 3072 1 π 4 ≈ 16 1 π 2 + 1536 1 π 4 ■
(d)
∫ 0 1 10 π ln sec x d x ≈ ∫ 0 1 10 π 1 2 x 2 + 1 12 x 4 d x = [ 1 6 x 3 + 1 60 x 5 ] 0 1 10 π = 0.005219 ( 4 sf ) ■ \begin{align*}
& \int_0^{\frac{1}{10}\pi} \ln \sec x \, \mathrm{d}x \\
& \approx \int_0^{\frac{1}{10}\pi} \frac{1}{2} x^2 + \frac{1}{12} x^4 \, \mathrm{d}x \\
&= \left[ \frac{1}{6} x^3 + \frac{1}{60} x^5 \right]_0^{\frac{1}{10}\pi} \\
&= 0.005219 \; (\textrm{4 sf}) \; \blacksquare
\end{align*} ∫ 0 10 1 π ln sec x d x ≈ ∫ 0 10 1 π 2 1 x 2 + 12 1 x 4 d x = [ 6 1 x 3 + 60 1 x 5 ] 0 10 1 π = 0.005219 ( 4 sf ) ■
Question Commentary A rather typical Maclaurin expansion question (via differentiation). We employed two tricks
to simplify our differentiation process.
First we used implicit differentiation
immediately by taking exponents on both sides straightaway. We also made use of the
tan 2 x + 1 = sec 2 x \tan^2 x + 1 = \sec^2 x tan 2 x + 1 = sec 2 x