Exercise [06.10]

Sameed Zahoor · 13 May 2008, 09:16

1. For

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.. $d(\tan{(\arctan{x})})=\sec^2{(\arctan{x})}d(\arctan{x})$
$\Rightarrow\ dx=(1+\tan^2{(\arctan{x})})d(\arctan{x})$
$\Rightarrow\ d(\arctan{x})=\frac{dx}{1+x^2}$

Exercise [06.10]

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