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<!DOCTYPE html>
<meta charset="utf-8">
<title>Successive Differentiation | Calculus Made Easy</title>
<link rel="stylesheet" href="screen.css">
<style>
body{counter-reset:h1 7}
</style>
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<h1><br>Successive Differentiation</h1>
<p class="a rotatedFloralHeartBullet">
<p>Let us try the effect of repeating several times over
the operation of differentiating a function (see <a href="3.html#function">here</a>).
Begin with a concrete case.
<p>Let $y = x^5$.
\begin{alignat*}{3}
&\text{First differentiation, } &&5x^4. && \\
&\text{Second differentiation, } &&5 × 4x^3 &&= 20x^3. \\
&\text{Third differentiation, } &&5 × 4 × 3x^2 &&= 60x^2. \\
&\text{Fourth differentiation, } &&5 × 4 × 3 × 2x &&= 120x. \\
&\text{Fifth differentiation, } &&5 × 4 × 3 × 2 × 1 &&= 120. \\
&\text{Sixth differentiation, } && &&= 0.
\end{alignat*}
<p>There is a certain notation, with which we are
already acquainted (see <a href="3.html#notation">here</a>), 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 $ax^n$, or $\cos x$ or any other complicated
function of $x$.
<p>The corresponding symbol for the differential coefficient
is $f'(x)$, which is simpler to write than $\dfrac{dy}{dx}$.
This is called the “derived function” of $x$.
<p>Suppose we differentiate over again, we shall get
the “second derived function” or second differential
coefficient, which is denoted by $f''(x)$; and so on.
<p>Now let us generalize.
<p>Let $y = f(x) = x^n$.
\begin{align*}
\text{First differentiation,}\; f'(x) &= nx^{n-1}. \\
\text{Second differentiation,}\; f''(x) &= n(n-1)x^{n-2}. \\
\text{Third differentiation,}\; f'''(x) &= n(n-1)(n-2)x^{n-3}. \\
\text{Fourth differentiation,}\; f''''(x) &= n(n-1)(n-2)(n-3)x^{n-4}. \\
&\llap{\text{etc.,}} \text{ etc.}
\end{align*}
<p>But this is not the only way of indicating successive
differentiations. For,
\begin{align*}
\text{if the original function be}\; y &= f(x); \\
\text{once differentiating gives}\; \frac{dy}{dx} &= f'(x); \\
\text{twice differentiating gives}\; \frac{d\left(\dfrac{dy}{dx}\right)}{dx} &= f''(x);
\end{align*}
and this is more conveniently written as $\dfrac{d^2y}{(dx)^2}$, or
more usually $\dfrac{d^2y}{dx^2}$. Similarly, we may write as the
result of thrice differentiating, $\dfrac{d^3y}{dx^3} = f'''(x)$.
<hr>
<p>
<p><em>Examples</em>
Now let us try $y = f(x) = 7x^4 + 3.5x^3 - \frac{1}{2}x^2 + x - 2$.
\begin{align*}
\frac{dy}{dx} &= f'(x) = 28x^3 + 10.5x^2 - x + 1, \\
\frac{d^2y}{dx^2} &= f''(x) = 84x^2 + 21x - 1, \\
\frac{d^3y}{dx^3} &= f'''(x) = 168x + 21, \\
\frac{d^4y}{dx^4} &= f''''(x) = 168, \\
\frac{d^5y}{dx^5} &= f'''''(x) = 0.
\end{align*}
In a similar manner if $y = \phi(x) = 3x(x^2 - 4)$,
\begin{align*}
\phi'(x) &= \frac{dy}{dx} = 3\bigl[x × 2x + (x^2 - 4) × 1\bigr] = 3(3x^2 - 4), \\
\phi''(x) &= \frac{d^2y}{dx^2} = 3 × 6x = 18x, \\
\phi'''(x) &= \frac{d^3y}{dx^3} = 18, \\
\phi''''(x) &= \frac{d^4y}{dx^4} = 0.
\end{align*}
<p>
<hr><h3>Exercises IV</h3>
<p>Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ for the following expressions:
<p>(1) $y = 17x + 12x^2$.
<p>(2) $y = \dfrac{x^2 + a}{x + a}$.
<p>(3) $y = 1 + \dfrac{x}{1} + \dfrac{x^2}{1×2} + \dfrac{x^3}{1×2×3} + \dfrac{x^4}{1×2×3×4}$.
<p>(4) Find the 2nd and 3rd derived functions in
the Exercises III. (<a href="6.html#examples2">here</a>), No. 1 to No. 7, and in the
Examples given (<a href="6.html#examples3">here</a>), No. 1 to No. 7.
<p><h3 class="answers">Answers</h3>
<p>(1) $17 + 24x$; $24$.
<p>(2) $\dfrac{x^2 + 2ax - a}{(x + a)^2}$; $\dfrac{2a(a + 1)}{(x + a)^3}$.
<p>(3) $1 + x + \dfrac{x^2}{1 × 2} + \dfrac{x^3}{1 × 2 × 3}$; $1 + x + \dfrac{x^2}{1 × 2}$.
<p><em>Exercises III</em>
<p>(4) (<em>Exercises III.</em> ):
<p>(1) (<em>a</em> ) $\dfrac{d^2 y}{dx^2} = \dfrac{d^3 y}{dx^3} = 1 + x + \frac{1}{2}x^2 + \frac{1}{6} x^3 + \ldots$.
<p>(<em>b</em> ) $2a$, $0$.
<p>(<em>c</em> ) $2$, $0$.
<p>(<em>d</em> ) $6x + 6a$, $6$.
<p>(2) $-b$, $0$.
<p>(3) $2$, $0$.
<p>(4) $\begin{gathered}[t]
56440x^3 - 196212x^2 - 4488x + 8192. \\
169320x^2 - 392424x - 4488.
\end{gathered}$
<p>(5) $2$, $0$.
<p>(6) $371.80453x$, $371.80453$.
<p>(7) $\dfrac{30}{(3x + 2)^3}$, $-\dfrac{270}{(3x + 2)^4}$.
<p><em>Examples</em> ):
<p>(1) $\dfrac{6a}{b^2} x$, $\dfrac{6a}{b^2}$.
<p>(2) $\dfrac{3a \sqrt{b}} {2 \sqrt{x}} - \dfrac{6b \sqrt[3]{a}}{x^3}$,
$\dfrac{18b \sqrt[3]{a}}{x^4} - \dfrac{3a \sqrt{b}}{4 \sqrt{x^3}}$.
<p>(3)
$\dfrac{2}{\sqrt[3]{\theta^8}} - \dfrac{1.056}{\sqrt[5]{\theta^{11}}}$,
$\dfrac{2.3232}{\sqrt[5]{\theta^{16}}} - \dfrac{16}{3 \sqrt[3]{\theta^{11}}}$.
<p>(4) $\begin{gathered}[t]
810t^4 - 648t^3 + 479.52t^2 - 139.968t + 26.64. \\
3240t^3 - 1944t^2 + 959.04t - 139.968.
\end{gathered}$
<p>(5) $12x + 2$, $12$.
<p>(6) $6x^2 - 9x$, $12x - 9$.
<p>(7)
$\begin{aligned}[t]
&\dfrac{3}{4} \left(\dfrac{1}{\sqrt{\theta}} + \dfrac{1}{\sqrt{\theta^5}}\right)
+\dfrac{1}{4} \left(\dfrac{15}{\sqrt{\theta^7}} - \dfrac{1}{\sqrt{\theta^3}}\right). \\
&\dfrac{3}{8} \left(\dfrac{1}{\sqrt{\theta^5}} - \dfrac{1}{\sqrt{\theta^3}}\right)
-\dfrac{15}{8}\left(\dfrac{7}{\sqrt{\theta^9}} + \dfrac{1}{\sqrt{\theta^7}}\right).
\end{aligned}$
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