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x = tan ( y ) {\displaystyle x=\tan \left(y\right)} 1 = sec 2 ( y ) ∗ d y d x {\displaystyle 1=\sec ^{2}\left(y\right)*{\frac {dy}{dx}}} (Chain rule, derivative of tan=sec^2) 1 sec 2 ( y ) = d y d x {\displaystyle {\frac {1}{\sec ^{2}\left(y\right)}}={\frac {dy}{dx}}} cos 2 ( y ) = d y d x {\displaystyle \cos ^{2}\left(y\right)={\frac {dy}{dx}}} d y d x = cos 2 ( y ) {\displaystyle {\frac {dy}{dx}}=\cos ^{2}\left(y\right)}
x 2 y + x y 2 = 6 {\displaystyle x^{2}y+xy^{2}=6\,} ( 2 x ∗ y + x 2 ∗ d y d x ) + ( 1 ∗ y 2 + x ∗ 2 y d y d x ) = 0 {\displaystyle \left(2x*y+x^{2}*{\frac {dy}{dx}}\right)+\left(1*y^{2}+x*2y{\frac {dy}{dx}}\right)=0} 2 x y + x 2 d y d x + y 2 + 2 x y d y d x = 0 {\displaystyle 2xy+x^{2}{\frac {dy}{dx}}+y^{2}+2xy{\frac {dy}{dx}}=0} x 2 d y d x + 2 x y d y d x = − 2 x y − y 2 {\displaystyle x^{2}{\frac {dy}{dx}}+2xy{\frac {dy}{dx}}=-2xy-y^{2}} d y d x = − 2 x y − y 2 x 2 + 2 x y {\displaystyle {\frac {dy}{dx}}={\frac {-2xy-y^{2}}{x^{2}+2xy}}} d y d x = − 2 x y + y 2 x 2 + 2 x y {\displaystyle {\frac {dy}{dx}}=-{\frac {2xy+y^{2}}{x^{2}+2xy}}}
To Find dy/dx for y = 2 cos ( ( 5 x ) 2 ) {\displaystyle y=2\cos \left(\left(5x\right)^{2}\right)}
you'll make 3 u's Let u = 2 cos ( u ) {\displaystyle {\text{Let }}u=2\cos \left(u\right)} Let u = u 2 {\displaystyle {\text{Let }}u=u^{2}\,} Let u = 5 x {\displaystyle {\text{Let }}u=5x\,}
Find d y d x {\displaystyle {\frac {dy}{dx}}\,} then find d 2 y d x 2 {\displaystyle {\frac {d^{2}y}{dx^{2}}}\,}
x 2 + y 2 = 1 {\displaystyle x^{2}+y^{2}=1\,} 2 x + 2 y d y d x = 0 {\displaystyle 2x+2y{\frac {dy}{dx}}=0\,}
d y d x = − 2 x 2 y {\displaystyle {\frac {dy}{dx}}={\frac {-2x}{2y}}\,} d y d x = − x y {\displaystyle {\frac {dy}{dx}}=-{\frac {x}{y}}\,}
2 + ( 2 d y d x ∗ d y d x + 2 y ∗ d 2 y d x 2 ) = 0 {\displaystyle 2+\left(2{\frac {dy}{dx}}*{\frac {dy}{dx}}+2y*{\frac {d^{2}y}{dx^{2}}}\right)=0\,} 2 ( d y d x ) 2 + 2 y d 2 y d x 2 = − 2 {\displaystyle 2\left({\frac {dy}{dx}}\right)^{2}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,} 2 ( − x y ) 2 + 2 y d 2 y d x 2 = − 2 {\displaystyle 2\left(-{\frac {x}{y}}\right)^{2}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,} 2 x 2 y 2 + 2 y d 2 y d x 2 = − 2 {\displaystyle 2{\frac {x^{2}}{y^{2}}}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,} 2 y d 2 y d x 2 = − 2 − 2 x 2 y 2 {\displaystyle 2y{\frac {d^{2}y}{dx^{2}}}=-2-2{\frac {x^{2}}{y^{2}}}\,} d 2 y d x 2 = − 2 − 2 x 2 y 2 2 y {\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {-2-2{\frac {x^{2}}{y^{2}}}}{2y}}\,} d 2 y d x 2 = − 1 y − x 2 y 3 {\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {1}{y}}-{\frac {x^{2}}{y^{3}}}\,} d 2 y d x 2 = − y 2 y 3 − x 2 y 3 {\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {y^{2}}{y^{3}}}-{\frac {x^{2}}{y^{3}}}\,} d 2 y d x 2 = − y 2 + x 2 y 3 {\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {y^{2}+x^{2}}{y^{3}}}\,} d 2 y d x 2 = − 1 y 3 {\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {1}{y^{3}}}\,}
x = 4 cos ( π 2 − t ⋅ π 60 ) {\displaystyle x=4\cos \left({\frac {\pi }{2}}-{\frac {t\cdot \pi }{60}}\right)} y = 4 sin ( π 2 − t ⋅ π 60 ) {\displaystyle y=4\sin \left({\frac {\pi }{2}}-{\frac {t\cdot \pi }{60}}\right)}
(Chain rule, derivative of tan=sec^2)
then find 
