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
If y = x5 + 3x3 − 2x + 7, then what are the higher derivatives?
The first derivative:
Now for the second derivative. We just differentiate our previous answer.
Now for the 3rd and 4th derivatives.
Now the 5th derivative is
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:
But we also know that velocity is the rate of change of displacement:
So it follows that the second derivative of displacement will give us acceleration:
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`.
s = 4t3 + 7t2 − 2t
At `t= 10`, the acceleration will be
`a=24(10) + 14 = 254` ms-2.
Easy to understand math videos:
Higher Derivatives of Implicit Functions
(The Answers for these two questions contain short video explanations.)
a. Find the second derivative of the implicit function xy + y2 = 4.
Now xy is a product, so we use Product Formula to obtain:
And we learned in the last section on Implicit Differentiation that
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)`
`[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.
We substitute our expression for `y'` from above, `dy/dx = (-y)/(x+2y)`:
Now multiply by `(x+2y)^2/(x+2y)^2` to give
Here's a movie giving a different view of this example. It uses:
- `dy/dx` notation;
- 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.
b. Find the value of the second derivative of the implicit function in part (a) when x = 2, where y > 0.
We need to find y when `x = 2`.
On substituting, we have:
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:
We substitute and obtain:
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.