# 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`.