# Parametric Differentiating

Sometimes we need to write derivatives with respect to x through differentials of another variable t. In this case expression for derivatives will be more complex.

So, let's calculate differentials with respect to t, in other words x is not an independent variable.

y'_x=(dy)/(dx)

y''_(x^2)=((dy)/(dx))'_x=(d((dy)/(dx)))/(dx)=((d^2y*dx-dy*d^2x)/(dx^2))/(dx)=(d^2y*dx-dy*d^2x)/(dx^3) (note that we used quotient rule)

y'''_(x^3)=((d^2y*dx-dy*d^2x)/(dx^3))'_x=(d((d^2y*dx-dy*d^2x)/(dx^3)))/(dx)=((dx^3(dx*d^3y-d^3x*dy)-3dx^2*d^2x(dx*d^2y-d^2x*dy))/(dx^6))/(dx) (note that we used product and quotient rule).

So, y'''_(x^3)=(dx(dx*d^3y-d^3x*dy)-3d^2x(dx*d^2y-d^2x*dy))/(dx^5).

These formulas allows us to perform parametric differentiation.

Suppose that we are given two functions x=u(t) and y=v(t), where t is independent variable.

When there exist derivatives of x and y with respect to t we can find derivatives of y with respect to x using above formulas.

Sometimes it is more convenient to express derivatives of y with respect to x in terms of derivatives (not differentials) of x and y with respect to t. We can easily obtain them from differential expressons, by dividing numerator and denominator by dt,dt^3,dt^5,... respectively.

Thus, y'_x=(dy)/(dx)=((dy)/(dt))/((dx)/(dt))=(y'_t)/(x'_t).

y''_(x^2)=(d^2y*dx-dy*d^2x)/(dx^3)=((d^2y)/(dt^2)*(dx)/(dt)-(dy)/(dt)*(d^2x)/(dt^2))/(((dx)/(dt))^3)=(x'_t y''_(t^2)-x''_(t^2) y'_t)/((x'_t)^3).

Similarly, we obtain expression for third derivative.

So,

color(green)(y'_x=(y'_t)/(x'_t)),

color(red)(y''_(x^2)=(x'_t y''_(t^2)-x''_(t^2) y'_t)/((x'_t)^3)),

color(blue)(y'''_(x^3)=(x'_t(x'_t y'''_(t^3)-x'''_(t^3)y'_t)-3x''_(t^2)(x'_t y''_(t^2)-x''_(t^2) y'_t))/((x'_t)^5)).

Example 1. Find (dy)/(dx) and (d^2y)/(dx^2) if y(t)=3t^3+5t and x(t)=2t^4+7t^2.

We have that y'_t=9t^2+5 and y''_(t^2)=18t. Now, x'_t=8t^3+14t and x''_(t^2)=24t^2+14.

So,

y'_x=(dy)/(dx)=(9t^2+5)/(8t^3+14t).

y''_(x^2)=(d^2y)/(dx^2)=((8t^3+14t)18t-(24t^2+14)(9t^2+5))/((8t^3+14t)^3).

Note, that formulas for second derivative and third are fairly complex.

So, it is better to derive these formulas while solving the problem.

Example 2. Find (d^2y)/(dx^2) if y(t)=t^2+1 and x(t)=t^2+t.

Let's first find first derivative:

y'_x=(y'_t)/(x'_t)=(2t)/(2t+1).

We can find second derivative using above formula, but we will take advantage of first derivative.

(d^2y)/(dx^2)=d/(dx) (dy)/(dx)=(d/(dt)((dy)/(dx)))/((dx)/(dt))=(d/(dt)((2t)/(2t+1)))/((dx)/(dt))=(((2t)'(2t+1)-2t(2t+1)')/(2t+1)^2)/(2t+1)=(4)/(2t+1)^3.