9. Higher Derivatives

by M. Bourne

We can continue to find the derivatives of a derivative. We find the

  • second derivative by taking the derivative of the first derivative,
  • third derivative by taking the derivative of the second derivative... etc

Example 1

If y = x5 + 3x3 − 2x + 7, then what are the higher derivatives?

Answer

The first derivative:

`(dy)/(dx)=y'=5x^4+9x^2-2`

Now for the second derivative. We just differentiate our previous answer.

`(d^2y)/(dx^2)=y''=20x^3+18x`

Now for the 3rd and 4th derivatives.

`(d^3y)/(dx^3)=y'''=60x^2+18`

`(d^4y)/(dx^4)=y^("iv")=120x`

Now the 5th derivative is

`y^"v"=120`

The 6th, 7th, 8th and all other derivatives are 0, since the derivative of a constant is 0.

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Application - Acceleration

We saw before that acceleration is the rate of change of velocity:

`a=(dv)/(dt)`

But we also know that velocity is the rate of change of displacement:

`v=(ds)/(dt)`

So it follows that the second derivative of displacement will give us acceleration:

`a=(d^2s)/(dt^2)`

Example 2

If the displacement (in metres) at time t (in seconds) of an object is given by

s = 4t3 + 7t2 − 2t,

find the acceleration at time `t = 10`.

Answer

s = 4t3 + 7t2 − 2t

`v=(ds)/(dt)=12t^2+14t-2`

`a=(d^2s)/(dt^2)=24t+14`

At `t= 10`, the acceleration will be

`a=24(10) + 14 = 254` ms-2.

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Higher Derivatives of Implicit Functions

Example 3

(The Answers for these two questions contain short video explanations.)

a. Find the second derivative of the implicit function xy + y2 = 4.

Answer

First derivative:

Now xy is a product, so we use Product Formula to obtain:

`d/dx(xy)=xy'+y`

And we learned in the last section on Implicit Differentiation that

`d/(dx)y^2=2y(dy)/(dx)`

We can write this as:

`d/dx(y^2) = 2yy'`

Putting it together, here is the first derivative of our implicit function:

`xy' + y + 2yy' = 0`

[I am using `y'` instead of `dy/dx`. It is easier to type and quite easy to read.]

We could write this as: `dy/dx = -y/(x+2y)`

Second derivative:

`[xy'' + y'] + [ y'] +` ` [2yy'' + y'(2y')] ` `= 0`

and this simplifies to:

`(x + 2y)y'' + 2y' + 2(y')^2 = 0`

We can solve for `y''`:

`y'' = [-2(y' + (y')^2)] / (x + 2y)`

We could leave our answer as is, or simplify it even further by re-expressing it in terms of `x` and `y` only, as follows.

Further simplification

We substitute our expression for `y'` from above, `dy/dx = (-y)/(x+2y)`:

`(d^2y)/dx^2=(-2 (dy/dx+(dy/dx)^2))/(x+2y)

`=-2(((-y)/(x+2y))+((-y)/(x+2y))^2)/(x+2y)`

Now multiply by `(x+2y)^2/(x+2y)^2` to give

`=-2((-y(x+2y)+(y)^2)/(x+2y)^3)`

`=-2((-yx-2y^2+y^2)/(x+2y)^3)`

`=-2((-yx-y^2)/(x+2y)^3)`

`=(2y(x+y))/((x+2y)^3)`

Video

Here's a movie giving a different view of this example. It uses:

  1. `dy/dx` notation;
  2. A different approach to the problem (in the movie, we find the expression for `dy/dx` first, then differentiate that to get the second derivative). The result is a simpler expression for the second derivative.

The answer looks quite different, but it does have the same value.

Easy to understand math videos:
MathTutorDVD.com

b. Find the value of the second derivative of the implicit function in part (a) when x = 2, where y > 0.

Answer

We need to find y when `x = 2`.

On substituting, we have:

`2(y)+y^2=4`

Solving for `y` using the quadratic formula gives:

`y^2 +2y - 4=0`

`y=(-2+- sqrt(2^2 + 16))/2`

`y=-1 +- sqrt(5)`

Now, since the question says `y > 0`, we choose `y=-1+sqrt(5)=1.236` only.

Using the result from the first part of the question:

`(d^2y)/(dx^2)=(2y(x+y))/((x+2y)^3)`

We substitute and obtain:

`=(2(1.236)(2+1.236))/((2+2(1.236))^3)`

`=(2(1.236)(2-3(1.236)))/((2+2(1.236)^3)`

`=0.0894`

Video

Here is a movie of the solution. The form of the second derivative here is different to the above solution, but is also quite correct.

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