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Find <math>\frac{dy}{dx}\,</math> then find <math>\frac{d^2y}{dx^2}\,</math> <br /><br />
Find <math>\frac{dy}{dx}\,</math> then find <math>\frac{d^2y}{dx^2}\,</math> <br /><br />
<math>x^{2} + y^{2} = 1\,</math><br /><br />
<math>x^2 + y^2 = 1\,</math><br /><br />
<math>2x + 2y\frac{dy}{dx} = 0\,</math><br /><br />
===Find first derivative===
<math>\frac{dy}{dx} = \frac{-2x}{2y}\,</math><br /><br />
<math>\frac{dy}{dx} = -\frac{x}{y}\,</math><br /><br />
===Find second derivative===
<math>2 + \left(2\frac{dy}{dx}*\frac{dy}{dx} + 2y*\frac{d^{2}y}{dx^2}\right) = 0\,</math><br /><br />
<math>2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br />
<math>2\left(-\frac{x}{y}\right)^2 + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br />
<math>2\frac{x^2}{y^2} + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br />
<math>2y\frac{d^{2}y}{dx^2} = -2-2\frac{x^2}{y^2}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = \frac{-2-2\frac{x^2}{y^2}}{2y}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = -\frac{1}{y} - \frac{x^2}{y^3}\,</math><br /><br />
Revision as of 23:11, 23 October 2008
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}}}
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)}
9~
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}}}
Multiple u's
To Find dy/dx for
y
=
2
cos
(
(
5
x
)
2
)
{\displaystyle y=2\cos \left(\left(5x\right)^{2}\right)}
The way she explains it
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\,}
Gaaah, help
Find
d
y
d
x
{\displaystyle {\frac {dy}{dx}}\,}
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\,}
Find first derivative
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}}\,}
Find second derivative
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}}}\,}
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↑
(Chain rule, derivative of tan=sec^2)
then find 
