Similarly for higher derivatives. This transformation is called changing the dependent variable from y to z , the independent variable remaining x throughout. We will now illustrate this process by means of an example.
Illustrative Example 1.
(E )
d
2
y
d
x
2
=
1
+
2
(
1
+
y
)
1
+
y
2
(
d
y
d
x
)
2
,
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=1+{\frac {2(1+y)}{1+y^{2}}}\left({\frac {dy}{dx}}\right)^{2},}
change the dependent variable from y to z by means of the relation
(F )
y
=
tan
z
.
{\displaystyle y=\tan z.}
Solution. From (F ) ,
d
y
d
x
=
sec
2
z
d
z
d
x
{\displaystyle {\frac {dy}{dx}}=\sec ^{2}z{\frac {dz}{dx}}}
d
2
y
d
x
2
=
sec
2
d
2
z
d
x
2
+
2
sec
2
z
tan
z
(
d
z
d
x
)
2
,
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=\sec ^{2}{\frac {d^{2}z}{dx^{2}}}+2\sec ^{2}z\tan z\left({\frac {dz}{dx}}\right)^{2},}
Substituting in (E ) ,
sec
2
z
d
2
z
d
x
2
+
2
sec
2
z
tan
z
(
d
z
d
x
)
2
=
1
2
(
1
+
tan
z
)
1
+
tan
2
z
(
sec
2
z
d
z
d
x
)
2
,
{\displaystyle \sec ^{2}z{\frac {d^{2}z}{dx^{2}}}+2\sec ^{2}z\tan z\left({\frac {dz}{dx}}\right)^{2}=1{\frac {2(1+\tan z)}{1+\tan ^{2}z}}\left(\sec ^{2}z{\frac {dz}{dx}}\right)^{2},}
and reducing, we get
d
2
z
d
x
2
−
2
(
d
z
d
x
)
2
=
cos
2
z
{\displaystyle {\tfrac {d^{2}z}{dx^{2}}}-2\left({\tfrac {dz}{dx}}\right)^{2}=\cos ^{2}z}
Ans. 97. Change of the independent variable. y be a function of x , and at the same time let x (and hence also y ) be a function of a new variable t . It is required to express
d
y
d
x
,
d
2
y
d
x
2
{\displaystyle {\frac {dy}{dx}},\ {\frac {d^{2}y}{dx^{2}}}}
in terms of new derivatives having t as the independent variable.
By XXV , §33
d
y
d
t
{\displaystyle {\frac {dy}{dt}}}
=
d
y
d
x
d
x
d
t
{\displaystyle ={\frac {dy}{dx}}{\frac {dx}{dt}}}
(A )
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
=
d
y
d
t
d
x
d
t
{\displaystyle ={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}}
Also
d
2
y
d
x
2
=
d
d
x
(
d
y
d
x
)
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)}
=
d
d
t
(
d
y
d
x
)
d
t
d
x
=
d
d
t
(
d
y
d
x
)
d
x
d
t
{\displaystyle ={\frac {d}{dt}}\left({\frac {dy}{dx}}\right){\frac {dt}{dx}}={\frac {{\frac {d}{dt}}\left({\frac {dy}{dx}}\right)}{\frac {dx}{dt}}}}
But differentiating (A ) with respect to t ,
d
d
t
(
d
y
d
x
)
=
d
d
t
(
d
y
d
t
d
x
d
t
)
{\displaystyle {\frac {d}{dt}}\left({\frac {dy}{dx}}\right)={\frac {d}{dt}}\left({\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}\right)}
=
d
x
d
t
d
2
y
d
t
2
−
d
y
d
t
d
2
x
d
t
2
(
d
x
d
t
)
2
{\displaystyle ={\frac {{\frac {dx}{dt}}{\frac {d^{2}y}{dt^{2}}}-{\frac {dy}{dt}}{\frac {d^{2}x}{dt^{2}}}}{\left({\frac {dx}{dt}}\right)^{2}}}}
