< User talk:Hayson1991 Browse history interactively ← Previous edit Next edit → Content deleted Content addedVisual Wikitext Inline Revision as of 23:02, 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\,}
The way you should do it
Let
y
=
2
cos
(
u
)
{\displaystyle {\text{Let }}y=2\cos \left(u\right)\,}
Let
u
=
v
2
{\displaystyle {\text{Let }}u=v^{2}\,}
Let
v
=
5
x
{\displaystyle {\text{Let }}v=5x\,}
d
y
d
x
=
d
y
d
u
∗
d
u
d
v
∗
d
v
d
x
{\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}*{\frac {du}{dv}}*{\frac {dv}{dx}}}
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\,}
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↑
(Chain rule, derivative of tan=sec^2)
then find 
