Simplest Cases

Now let us see how, on first principles, we can differentiate some simple algebraical expression.

Let $x$, then, grow a little bit bigger and become $x + dx$; similarly, $y$ will grow a bit bigger and will become $y + dy$. Then, clearly, it will still be true that the enlarged $y$ will be equal to the square of the enlarged $x$. Writing this down, we have: \begin{align*} y + dy &= (x + dx)^2.\\ \text{Doing the squaring we get:}\;\\ y + dy &= x^2 + 2x · dx+(dx)^2. \end{align*}

What does $(dx)^2$ mean? Remember that $dx$ meant a bit–a little bit–of $x$. Then $(dx)^2$ will mean a little bit of a little bit of $x$; that is, as explained above (here), it is a small quantity of the second order of smallness. It may therefore be discarded as quite inconsiderable in comparison with the other terms. Leaving it out, we then have: \begin{align*} y + dy &= x^2 + 2x · dx. \\ \text{Now $y=x^2$; so let us subtract this from the equation and we have left}\;\\ dy &= 2x · dx. \\ \text{ Dividing across by $dx$, we find}\;\\ \frac{dy}{dx} &= 2x. \end{align*}

*Note*–This ratio $\dfrac{dy}{dx}$ is the result of differentiating $y$ with
respect to $x$. Differentiating means finding the differential coefficient.
Suppose we had some other function of $x$, as, for
example, $u = 7x^2 + 3$. Then if we were told to differentiate this
with respect to $x$, we should have to find $\dfrac{du}{dx}$, or, what is the same
thing, $\dfrac{d(7x^2 + 3)}{dx}$. On the other hand, we may have a case in which
time was the independent variable (see

*Numerical example.*

Suppose $x=100$ and $\therefore y=10,000$. Then let $x$ grow till it becomes $101$ (that is, let $dx=1$). Then the enlarged $y$ will be $101 × 101 = 10,201$. But if we agree that we may ignore small quantities of the second order, $1$ may be rejected as compared with $10,000$; so we may round off the enlarged $y$ to $10,200$. $y$ has grown from $10,000$ to $10,200$; the bit added on is $dy$, which is therefore $200$.

$\dfrac{dy}{dx} = \dfrac{200}{1} = 200$. According to the algebra-working of the previous paragraph, we find $\dfrac{dy}{dx} = 2x$. And so it is; for $x=100$ and $2x=200$.

But, you will say, we neglected a whole unit.

Well, try again, making $dx$ a still smaller bit.

Try $dx=\frac{1}{10}$. Then $x+dx=100.1$, and \[ (x+dx)^2 = 100.1 × 100.1 = 10,020.01. \]

Now the last figure $1$ is only one-millionth part of the $10,000$, and is utterly negligible; so we may take $10,020$ without the little decimal at the end. And this makes $dy=20$; and $\dfrac{dy}{dx} = \dfrac{20}{0.1} = 200$, which is still the same as $2x$.

We let $y$ grow to $y+dy$, while $x$ grows to $x+dx$.

Then we have \[ y + dy = (x + dx)^3. \]

Doing the cubing we obtain \[ y + dy = x^3 + 3x^2 · dx + 3x(dx)^2+(dx)^3. \]

Now we know that we may neglect small quantities of the second and third orders; since, when $dy$ and $dx$ are both made indefinitely small, $(dx)^2$ and $(dx)^3$ will become indefinitely smaller by comparison. So, regarding them as negligible, we have left: \[ y + dy=x^3+3x^2 · dx. \]

But $y=x^3$; and, subtracting this, we have: \begin{align*} dy &= 3x^2 · dx, \\ \text{ and}\; \frac{dy}{dx} &= 3x^2. \end{align*}

Now all these cases are quite easy. Let us collect the results to see if we can infer any general rule. Put them in two columns, the values of $y$ in one and the corresponding values found for $\dfrac{dy}{dx}$ in the other: thus

$y$ | $\frac{dy}{dx}$ |
---|---|

$x^2$ | $2x$ |

$x^3$ | $3x^2$ |

$x^4$ | $4x^3$ |

Following out logically our observation, we should conclude that if we want to deal with any higher power,–call it $n$–we could tackle it in the same way.

Let $y = x^n,$

then, we should expect to find that

$\frac{dy}{dx} = nx^{(n-1)}$.

For example, let $n=8$, then $y=x^8$; and differentiating it would give $\dfrac{dy}{dx} = 8x^7$.

And, indeed, the rule that differentiating $x^n$ gives as the result $nx^{n-1}$ is true for all cases where $n$ is a whole number and positive. [Expanding $(x + dx)^n$ by the binomial theorem will at once show this.] But the question whether it is true for cases where $n$ has negative or fractional values requires further consideration.

*Case of a negative power.*

Let $y = x^{-2}$. Then proceed as before: \begin{align*} y+dy &= (x+dx)^{-2} \\ &= x^{-2} \left(1 + \frac{dx}{x}\right)^{-2}. \end{align*} Expanding this by the binomial theorem (see here), we get \begin{align*} &=x^{-2} \left[1 - \frac{2\, dx}{x} + \frac{2(2+1)}{1×2} \left(\frac{dx}{x}\right)^2 - \text{etc.}\right] \\ &=x^{-2} - 2x^{-3} · dx + 3x^{-4}(dx)^2 - 4x^{-5}(dx)^3 + \text{etc.} \\ \end{align*} So, neglecting the small quantities of higher orders of smallness, we have: \begin{align*} y + dy &= x^{-2} - 2x^{-3} · dx. \end{align*} Subtracting the original $y = x^{-2}$, we find \begin{align*} dy &= -2x^{-3}dx, \\ \frac{dy}{dx} &= -2x^{-3}. \end{align*} And this is still in accordance with the rule inferred above.

*Case of a fractional power.*

Let $y= x^{\frac{1}{2}}$. Then, as before, \begin{align*} y+dy &= (x+dx)^{\frac{1}{2}} = x^{\frac{1}{2}} (1 + \frac{dx}{x} )^{\frac{1}{2}} \\ &= \sqrt{x} + \frac{1}{2} \frac{dx}{\sqrt{x}} - \frac{1}{8} \frac{(dx)^2}{x\sqrt{x}} + \text{terms with higher powers of $dx$.} \end{align*}

Subtracting the original $y = x^{\frac{1}{2}}$, and neglecting higher powers we have left: \[ dy = \frac{1}{2} \frac{dx}{\sqrt{x}} = \frac{1}{2} x^{-\frac{1}{2}} · dx, \] and $\dfrac{dy}{dx} = \dfrac{1}{2} x^{-\frac{1}{2}}$. Agreeing with the general rule.

*Summary.* Let us see how far we have got. We
have arrived at the following rule: To differentiate $x^n$,
multiply by the power and reduce the power by
one, so giving us $nx^{n-1}$ as the result.

Differentiate the following:

(1) $y = x^{13}$

(2) $y = x^{-\frac{3}{2}}$

(3) $y = x^{2a}$

(4) $u = t^{2.4}$

(5) $z = \sqrt[3]{u}$

(6) $y = \sqrt[3]{x^{-5}}$

(7) $u = \sqrt[5]{\dfrac{1}{x^8}}$

(8) $y = 2x^a$

(9) $y = \sqrt[q]{x^3}$

(10) $y = \sqrt[n]{\dfrac{1}{x^m}}$

*You have now learned how to differentiate powers
of $x$. How easy it is!*

(1) $\dfrac{dy}{dx} = 13x^{12}$.

(2) $\dfrac{dy}{dx} = - \dfrac{3}{2} x^{-\frac{5}{2}}$.

(3) $\dfrac{dy}{dx} = 2ax^{(2a-1)}$.

(4) $\dfrac{du}{dt} = 2.4t^{1.4}$.

(5) $\dfrac{dz}{du} = \dfrac{1}{3} u^{-\frac{2}{3}}$.

(6) $\dfrac{dy}{dx} = -\dfrac{5}{3}x^{-\frac{8}{3}}$.

(7) $\dfrac{du}{dx} = -\dfrac{8}{5}x^{-\frac{13}{5}}$.

(8) $\dfrac{dy}{dx} = 2ax^{a-1}$.

(9) $\dfrac{dy}{dx} = \dfrac{3}{q} x^{\frac{3-q}{q}}$.

(10) $\dfrac{dy}{dx} = -\dfrac{m}{n} x^{-\frac{m+n}{n}}$.

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