5.7 Higher-Order Derivatives
Now that we mastered using different rules to find derivatives, we have one more thing to learn: higher-order derivatives. As you may have noticed, the derivative of a function is a function. Therefore, we can differentiate a derivative. A derivative of the derivative of a function is called the second derivative because we differentiate it twice. It has some useful features.
The second derivative is often denoted as f′′(x),y′′, or d2ydx2 . The second derivative is used to find the acceleration at a certain value of x and the curvature of the graph of a function, which we will discuss in the next chapter. Putting the application aside for now, let’s practice finding the second derivative!
Example 1
What is the second derivative of the following equation?
f(x)=x5−3x2−4x+1
First, we find the first derivative, which is
f′(x)=5x4−6x−4
Therefore, the second derivative is
f′′(x)=20x3−6
Example 2
Let’s practice one more:
f(x)=ex5−3x2−4x−1
Recall that the derivative of eg(x) is
g′(x)eg(x)
where
g(x)=x5−3x2−4x−1
Since
g′(x)=5x4−6x−4
the first derivative of f(x) is
f′(x)=(5x4−6x−4)ex5−3x2−4x−1
For the second derivative, we have to use the product rule. Let
g(x)=5x4−6x−4f(x)=ex5−3x2−4x−1
since
f′(x)=g(x)f(x)
The derivatives of g(x) and f(x) are
g′(x)=20x3−6f′(x)=(5x4−6x−4)ex5−3x2−4x−1
Therefore, by applying the product rule, we get
f"(x)=g′(x)f(x)+g(x)f′(x)=(20x3−6)ex5−3x2−4x−1+(5x4−6x−4)(5x4−6x−4)ex5−3x2−4x−1=[20x3−6+(5x4−6x−4)2]ex5−3x2−4x−1
In the above function, we just factored by the exponent term. You can further clean up the function within the brackets if you wish to.
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