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 Exercise [06.10] 
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Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Exercise [06.10]
1. For d(e^x) see soution to Exercise [6.5].
2. We know
d(e^{log{x}})=e^{log{x}}d(log{x})
or,d(x)=xd(log{x})
or,d(log{x})=\frac{dx}{x}
3.d(e^{ix})=ie^{ix}dx
\Rightarrow\ d(\cos{x}+i\sin{x})=i(\cos{x}+i\sin{x})dx
\Rightarrow\ d(\cos{x})+id(\sin{x})=-\sin{x}dx+i\cos{x}dx
Equating real and imaginary parts we get
d(\cos{x})=-\sin{x}dx
d(\sin{x})=\cos{x}dx
4.d(\tan{x})=d(\frac{\sin{x}}{\cos{x}})
=(\frac{\cos{x}d(\sin{x})-\sin{x}d(\cos{x})}{\cos^2{x}})dx
=(\frac{\cos^2{x}+\sin^2{x}}{\cos^2{x}})dx
=\frac{dx}{\cos^2{x}}
5.d(\sin{(\arcsin{x})})=\cos{(\arcsin{x})}d(\arcsin{x})
\Rightarrow\ dx=\sqrt{1-\sin^2{(\arcsin{x})}}d(\arcsin{x})
\Rightarrow\ d(\arcsin{x})=\frac{dx}{(1-x^2)^{1/2}}
6.d(\cos{(\arccos{x})})=-\sin{(\arccos{x})}d(\arccos{x})
\Rightarrow\ dx=-\sqrt{1-\cos^2{(\arccos{x})}}d(\arccos{x})
\Rightarrow\ d(\arccos{x})=-\frac{dx}{(1-x^2)^{1/2}}
7..d(\tan{(\arctan{x})})=\sec^2{(\arctan{x})}d(\arctan{x})
\Rightarrow\ dx=(1+\tan^2{(\arctan{x})})d(\arctan{x})
\Rightarrow\ d(\arctan{x})=\frac{dx}{1+x^2}


Last edited by Sameed Zahoor on 31 May 2008, 04:38, edited 10 times in total.

13 May 2008, 09:16

Joined: 11 May 2008, 20:07
Posts: 11
Location: france
Post Re: Exercise [6.10]
Hello Sameed,
you have forgotten a "d" before the log on your 4th line


13 May 2008, 10:00

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Exercise [6.10]
Thanks for pointing out.


13 May 2008, 10:05

Joined: 26 May 2008, 03:11
Posts: 1
Location: Atlanta, Ga.
Post Re: Exercise [6.10]
I believe the denominator is square rooted in the final solution for 5 and 6.


26 May 2008, 03:17
WWW

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Exercise [6.10]
Sorry,my mistake.


26 May 2008, 09:27
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