Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/173

giving ${\tfrac {d^{2}y}{dx^{2}}}$ in terms of ${\tfrac {dx}{dy}}$ and ${\tfrac {d^{2}x}{dy^{2}}}$. Similarly,

(37)

${\frac {d^{3}y}{dx^{3}}}=-{\frac {{\frac {d^{3}x}{dy^{3}}}{\frac {dx}{dy}}-3\left({\frac {d^{2}x}{dy^{2}}}\right)^{2}}{\left({\frac {dx}{dy}}\right)}}$;

and so on for higher derivatives. This transformation is called changing the independent variable from x to y.

Illustrative Example 1. Change the independent variable from x to y in the equation

$3\left({\frac {d^{2}y}{dx^{2}}}\right)^{2}-{\frac {dy}{dx}}{\frac {d^{3}y}{dx^{3}}}-{\frac {d^{2}y}{dx^{2}}}\left({\frac {dy}{dx}}\right)^{2}=0.$

Solution. Substituting from (35), (36), (37),

$3\left(-{\frac {\frac {d^{2}x}{dy^{2}}}{\left({\frac {dx}{dy}}\right)^{3}}}\right)^{2}-\left({\frac {1}{\frac {dx}{dy}}}\right)\left(-{\frac {{\frac {d^{3}x}{dy^{3}}}{\frac {dx}{dy}}-3\left({\frac {d^{2}x}{dy^{2}}}\right)^{2}}{\left({\frac {dx}{dy}}\right)^{5}}}\right)-\left(-{\frac {\frac {d^{2}x}{dy^{2}}}{\left({\frac {dx}{dy}}\right)^{3}}}\right)\left({\frac {1}{\frac {dx}{dy}}}\right)^{2}=0.$

Reducing, we get

${\frac {d^{3}x}{dy^{3}}}+{\frac {d^{2}x}{dy^{2}}}=0,$

a much simpler equation.

96. Change of the dependent variable. Let

(A)

$y=f(x),$

and suppose at the same time y is a function of z, say

(B)	$y=\psi (z).$

We may then express ${\tfrac {dy}{dx}}$, ${\tfrac {d^{2}y}{dx^{2}}}$ etc., in terms of ${\tfrac {dz}{dx}}$, ${\tfrac {d^{2}z}{dx^{2}}}$, etc., as follows

In general, z is a function of y by (B), §42; and since y is a function of x by (A), it is evident that z is a function of x. Hence by XXV we have

(C)	${\frac {dy}{dx}}={\frac {dy}{dz}}{\frac {dz}{dx}}=\psi '(z){\frac {dz}{dx}}$.
Also	${\tfrac {d^{2}y}{dx^{2}}}={\tfrac {d}{dx}}\left(\psi '(z){\tfrac {dz}{dx}}\right)={\tfrac {dz}{dx}}{\tfrac {d}{dx}}\psi '(z)+\psi '(z){\tfrac {d^{2}z}{dx^{2}}}$.	By V, §33
But	${\tfrac {d}{dx}}\psi '(z)={\tfrac {d}{dz}}\psi '(z){\tfrac {dz}{dx}}=\psi ''(z){\tfrac {dz}{dx}}.$	By XXV, §33
(D)	∴ ${\frac {d^{2}y}{dx^{2}}}=\psi ''(z)\left({\frac {dz}{dx}}\right)^{2}+\psi '(z){\frac {d^{2}z}{dx^{2}}}.$