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${\displaystyle x=\tan \left(y\right)}$

${\displaystyle 1=\sec ^{2}\left(y\right)*{\frac {dy}{dx}}}$ (Chain rule, derivative of tan=sec^2)

${\displaystyle {\frac {1}{\sec ^{2}\left(y\right)}}={\frac {dy}{dx}}}$

${\displaystyle \cos ^{2}\left(y\right)={\frac {dy}{dx}}}$

${\displaystyle {\frac {dy}{dx}}=\cos ^{2}\left(y\right)}$

9~

${\displaystyle x^{2}y+xy^{2}=6\,}$

${\displaystyle \left(2x*y+x^{2}*{\frac {dy}{dx}}\right)+\left(1*y^{2}+x*2y{\frac {dy}{dx}}\right)=0}$

${\displaystyle 2xy+x^{2}{\frac {dy}{dx}}+y^{2}+2xy{\frac {dy}{dx}}=0}$

${\displaystyle x^{2}{\frac {dy}{dx}}+2xy{\frac {dy}{dx}}=-2xy-y^{2}}$

${\displaystyle {\frac {dy}{dx}}={\frac {-2xy-y^{2}}{x^{2}+2xy}}}$

${\displaystyle {\frac {dy}{dx}}=-{\frac {2xy+y^{2}}{x^{2}+2xy}}}$

Multiple u's

To Find dy/dx for
${\displaystyle y=2\cos \left(\left(5x\right)^{2}\right)}$

The way she explains it

you'll make 3 u's
${\displaystyle {\text{Let }}u=2\cos \left(u\right)}$

${\displaystyle {\text{Let }}u=u^{2}\,}$

${\displaystyle {\text{Let }}u=5x\,}$

The way you should do it

${\displaystyle {\text{Let }}y=2\cos \left(u\right)\,}$

${\displaystyle {\text{Let }}u=v^{2}\,}$

${\displaystyle {\text{Let }}v=5x\,}$

${\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}*{\frac {du}{dv}}*{\frac {dv}{dx}}}$