7. Successive Differentiation

Let us try the effect of repeating several times over the operation of differentiating a function (see chapter 3). Begin with a concrete case.

Let $y=x^5$. $$\begin{array}{rlrlrl}& \text{First differentiation, }& & =5x^4.& & \\ & \text{Second differentiation, }& & 5\times 4x^3& & =20x^3.\\ & \text{Third differentiation, }& & 5\times 4\times 3x^2& & =60x^2.\\ & \text{Fourth differentiation, }& & 5\times 4\times 3\times 2x& & =120x.\\ & \text{Fifth differentiation, }& & 5\times 4\times 3\times 2\times 1& & =120.\\ & \text{Sixth differentiation, }& & & & =0.\end{array}$$

There is a certain notation, with which we are already acquainted (see chapter 3), used by some writers, that is very convenient. This is to employ the general symbol $f(x)$ for any function of $x$. Here the symbol $f()$ is read as “function of,” without saying what particular function is meant. So the statement $y=f(x)$ merely tells us that $y$ is a function of $x$, it may be $x^2$ or $a{x}^n$, or $\cos x$ or any other complicated function of $x$.

The corresponding symbol for the differential coefficient is $f^{\prime }(x)$, which is simpler to write than ${\displaystyle \frac{dy}{dx}}$. This is called the “derived function” of $x$.

Suppose we differentiate over again, we shall get the “second derived function” or second differential coefficient, which is denoted by $f^{\prime \prime }(x)$; and so on.

Now let us generalize.

Let $$y=f(x)=x^n$$.

$$\begin{array}{rlrl}& \text{First differentiation, }& & f^{\prime }(x)=n{x}^{n-1}.\\ & \text{Second differentiation, }& & f^{\prime \prime }(x)=n(n-1)x^{n-2}.\\ & \text{Third differentiation, }& & f^{\prime \prime \prime }(x)=5n(n-1)(n-2)x^{n-3}.\\ & \text{Fourth differentiation, }& & f^{\prime \prime \prime \prime }(x)=n(n-1)(n-2)(n-3)x^{n-4}.\\ & & & etc.\end{array}$$ 

But this is not the only way of indicating successive differentiations. For,

$$\begin{array}{rlrl}& \text{if the original function be}& & y=f(x)\\ & \text{once differentiating gives}& & {\displaystyle \frac{dy}{dx}}=f^{\prime }(x)\\ & \text{twice differentiating gives}& & {\displaystyle \frac{d{\displaystyle \frac{dy}{dx}}}{dx}}=f^{\prime \prime }(x)\end{array}$$

and this is more conveniently written as ${\displaystyle \frac{d^{2}y}{(dx{)}^2}}$, or more usually ${\displaystyle \frac{d^{2}y}{d{x}^2}}$. Similarly, we may write as the result of thrice differentiating, ${\displaystyle \frac{d^{3}y}{d{x}^3}}=f^{\prime \prime \prime }(x)$.

Examples.

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Exercises IV.

Find ${\displaystyle \frac{dy}{dx}}$ and ${\displaystyle \frac{d^{2}y}{d{x}^2}}$ for the following expressions:

$(1)y=17x+12x^2$

$(2)y={\displaystyle \frac{x^2+a}{x+a}}$

$(3)y=1+{\displaystyle \frac{x}{1}}+{\displaystyle \frac{x^2}{1\times 2}}+{\displaystyle \frac{x^3}{1\times 2\times 3}}+{\displaystyle \frac{x^4}{1\times 2\times 3\times 4}}$.

$(4)$ Find the 2nd and 3rd derived functions in the Exercises (chapter 6), No. 1 to No. 7, and in the Examples given (chapter 6), No. 1 to No. 7.