Higher order derivatives

Definition 1

Consider a function ff and its derivative f′f'. The derivative is often called the first derivative or first order derivative. As f′f' is again a function, we can determine its derivative as well. We call this new function the second derivative or second order derivative. It is denoted by f′′f^{\prime\prime} (say: ff double prime). The third derivative or third order derivative is the derivative of the second derivative, and so on:

f′ = derivative of ff′′ = derivative of f′f′′′ = derivative of f′′f(4) = derivative of f′′′f(5) = derivative of f(4)......\boxed{\begin{array}{ll} f^\prime & \text{ = derivative of $f$}\\ f^{\prime\prime} & \text{ = derivative of $f^{\prime}$}\\ f^{\prime\prime\prime} & \text{ = derivative of $f^{\prime\prime}$}\\ f^{(4)} & \text{ = derivative of $f^{\prime\prime\prime}$}\\ f^{(5)} & \text{ = derivative of $f^{(4)}$}\\ ... & ...\\ \end{array}}

Thus we have

Equation 1
f→′f′→′f′′→′f′′′→′f(4)→′…f \overset{\prime}{\rightarrow} f^\prime \overset{\prime}{\rightarrow} f^{\prime\prime} \overset{\prime}{\rightarrow} f^{\prime\prime\prime} \overset{\prime}{\rightarrow} f^{(4)} \overset{\prime}{\rightarrow} \dots

Higher order derivatives

Example 1

Determine the sixth order derivative of the function f(x)=x4f(x)=x^4.

Show

f′(x)=4x3f′′(x)=4⋅3x2=12x2f′′′(x)=12⋅2x1=24xf(4)(x)=24⋅1x0=24f(5)(x)=0f(6)(x)=0\begin{array}{lll} f'(x) & = 4 x^3 &\\ f^{\prime\prime}(x) & = 4 \cdot 3 x^2 & = 12 x^2 \\ f^{\prime\prime\prime}(x) & = 12\cdot 2 x^1 & = 24 x \\ f^{(4)}(x) & = 24\cdot 1x^0 & =24 \\ f^{(5)}(x) & = 0 & \\ f^{(6)}(x) & = 0 & \\ \end{array}

Note that in the example above, f′(x)=4x3f'(x)=4x^3 is the formula for calculating the slope of the tangents to the graph of ff. And similar, f′′(x)=12x2f^{\prime\prime}(x)=12x^2 is the formula for the slope of the tangents to the graph of f′f', and f′′′(x)=24xf^{\prime\prime\prime}(x)=24x is the formula for the slope of the tangents to the graph of f′′f^{\prime\prime}.

We can also find the higher order derivatives graphically. Given the graph of a function ff, we first sketch the graph f′f', then using this graph to sketch the graph of f′′f^{\prime\prime}, and so on.

Example 2

The graph of a function ff is shown below. Copy the graph (more or less), and sketch the graphs of f′f' and f′′f^{\prime\prime}.

Solution
Exercise 1

  1. Determine f(4)f^{(4)}:

    1. f(x)=x7f(x)=x^7
    2. g(x)=exg(x)=e^x
    3. h(x)=sin⁡(x)h(x)=\sin(x)
    4. k(x)=ln⁡(x)k(x)=\ln(x)
    5. m(x)=x⋅xm(x)=\sqrt{x}\cdot x

  2. Consider the function f(x)=3x4−5x3+2x2+1f(x)=3x^4-5x^3+2x^2+1. Determine f′′′(1)f^{\prime\prime\prime}(1).

Solution
  1. It is
    1. f(x)=x7→f′(x)=7x6→f′′(x)=7⋅6x5=42x5→f′′′(x)=42⋅5x4=210x4→f(4)(x)=210⋅4x3=840x3‾f(x)=x^7\rightarrow f^\prime(x)=7x^6\rightarrow f^{\prime\prime}(x)=7\cdot 6x^5=42x^5 \rightarrow f^{\prime\prime\prime}(x)=42\cdot 5x^4=210x^4\rightarrow f^{(4)}(x)=210\cdot 4x^3=\underline{840 x^3}
    2. g(x)=ex→g′(x)=g′′(x)=g′′′(x)=g(4)(x)=ex‾g(x)=e^x \rightarrow g^\prime(x)=g^{\prime\prime}(x)=g^{\prime\prime\prime}(x)=g^{(4)}(x)=\underline{e^x}
    3. h(x)=sin⁡(x)→h′(x)=cos⁡(x)→h′′(x)=−sin⁡(x)→h′′′(x)=−cos⁡(x)→h(4)(x)=−(−sin⁡(x))=sin⁡(x)‾h(x)=\sin(x) \rightarrow h^\prime(x)=\cos(x)\rightarrow h^{\prime\prime}(x)=-\sin(x) \rightarrow h^{\prime\prime\prime}(x)=-\cos(x) \rightarrow h^{(4)}(x)=-(-\sin(x))=\underline{\sin(x)} ... and it begins again!
    4. k(x)=ln⁡(x)→k′(x)=1x=x−1→k′′(x)=−1x−2=−x−2→k′′′(x)=−(−2x−3)=2x−3→k(4)(x)=2(−3x−4)=−6x−4=−6x4‾k(x)=\ln(x) \rightarrow k^\prime(x)=\frac{1}{x}=x^{-1}\rightarrow k^{\prime\prime}(x)=-1x^{-2}=-x^{-2} \rightarrow k^{\prime\prime\prime}(x)=-(-2x^{-3})=2x^{-3} \rightarrow k^{(4)}(x)=2(-3x^{-4})=-6x^{-4}=\underline{-\frac{6}{x^4}}
    5. m(x)=x⋅x=x1.5→m′(x)=1.5x0.5→m′′(x)=0.75x−0.5→m′′′(x)=−0.375x−1.5→m(4)=0.5625x−2.5‾m(x)=\sqrt{x}\cdot x = x^{1.5} \rightarrow m^\prime(x)=1.5x^{0.5} \rightarrow m^{\prime\prime}(x)=0.75 x^{-0.5} \rightarrow m^{\prime\prime\prime}(x)=-0.375x^{-1.5} \rightarrow m^{(4)}=\underline{0.5625 x^{-2.5}}
  2. f(x)=3x4−5x3+2x2+1→f′(x)=12x3−15x2+4x→f′′(x)=36x2−30x+4→f′′′(x)=72x−30→f′′′(1)=72⋅1−30=42‾f(x)=3x^4-5x^3+2x^2+1 \rightarrow f^\prime(x)=12x^3-15x^2+4x \rightarrow f^{\prime\prime}(x)=36x^2-30x+4 \rightarrow f^{\prime\prime\prime}(x)=72x-30 \rightarrow f^{\prime\prime\prime}(1)=72\cdot 1-30=\underline{42}