Kein Zurück: Looking for a good time?

$Differentiate\, \, \tan ^{3}(xy^{a}+y)=x$

Make the u substitution:
$\tan ^{3}u=x\;\; \; \; \;\; \; \; u=xy^{2}+y$

Differentiate:
$\frac{\mathrm{d} }{\mathrm{d} u}\left [ \right \tan ^{3}u]\frac{\mathrm{d} u}{\mathrm{d} x}=\frac{\mathrm{d} x}{\mathrm{d} x}\;\; \; \; \; \; \; \; \; \; \frac{\mathrm{d} u}{\mathrm{d} x}=y^{2}+2xy\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{\mathrm{d} y}{\mathrm{d} x}$

$(3\tan ^{2}u\sec ^{2}u)(y^{2}+2xy\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{\mathrm{d} y}{\mathrm{d} x})=1$

Now distribute, solve for dy/dx and simplify! :D

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-3y^{2}\tan ^{2}(xy^{2}+y)\sec ^{2}(xy^{2}+y)}{3(2xy+1)\tan ^{2}(xy^{2}+y)\sec ^{2}(xy^{2}+y)}$

Kein Zurück

Tuesday, June 23, 2009

Looking for a good time?

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