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,

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

Change the independent variable from x to y in the equation


 * 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


 * a much simpler equation.

Let

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

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