I have a question that is highly similar to this one: Differentiation from first principles of specific form. However, I'm not sure how to do this when fractional powers are involved. I understand that the answer involving the binomial theorem shows this for fractions too, but I'm not sure how to go through the mechanics of it all. For example, when trying to differentiate $\sqrt{ax+b}$, I get to

$$ \lim_{h \rightarrow 0} \frac{\sqrt{a(x+h)+b} - \sqrt{ax+b}}{h} $$

then can't get the $h$ to cancel out. I think this involves algebraic manipulation of some sort, but can't figure it out.


  • $\begingroup$ Hint: Multiple the numerator and denominator by $\sqrt{a(x+h)+b}+\sqrt{ax+b}$. $\endgroup$ – Julian Rosen May 6 '13 at 1:48
  • $\begingroup$ Right...why didn't I see that before..duh! Thank you! $\endgroup$ – Samuel Tan May 6 '13 at 1:50
  • $\begingroup$ How would this work for $(ax+b)^{1/3}$? $\endgroup$ – Samuel Tan May 6 '13 at 1:51
  • $\begingroup$ In general, for $(ax+b)^{1/n}$, you can multiply numerator and denominator by $\displaystyle \sum_{j=0}^{n-1} (a(x+h)+b)^{\frac{j}{n}}(ax+b)^{\frac{n-1-j}{n}}$. After multiplying out, the numerator will just be $ah$. $\endgroup$ – Julian Rosen May 6 '13 at 1:56

$$\dfrac{\sqrt{a(x+h)+b} - \sqrt{ax+b}}h = \dfrac{\sqrt{a(x+h)+b} - \sqrt{ax+b}}h \times \dfrac{\sqrt{a(x+h)+b} + \sqrt{ax+b}}{\sqrt{a(x+h)+b} + \sqrt{ax+b}}$$ Simplifying the above gives us $$\dfrac{a(x+h)+b - (ax+b)}{h\left(\sqrt{a(x+h)+b} + \sqrt{ax+b} \right)} = \dfrac{a}{\sqrt{a(x+h)+b} + \sqrt{ax+b}}$$ Now you should be able to finish it off.


Let $m$ and $n$ be positive integers. It does not matter whether $m$ and $n$ have common factors. I will show that the derivative of $x^{\frac{m}{n}}$ is $\frac{m}{n}x^{\frac{m}{n} - 1}$ using the limit definition of the derivative.

To this end, let $f(x) = x^{\frac{m}{n}}$. Then the $h$-difference quotient of the function $f$ is $$ \frac{f(x+h) \; - \; f(x)}{h} \;\;= \;\; \frac{(x+h)^{\frac{m}{n}} \; - \; x^{\frac{m}{n}}}{h} $$ I will work with this situation in essentially the same way that you work with $x^{\frac{1}{2}}$ and $x^{\frac{1}{3}}$, namely by rationalizing the numerator.

To rationalize a sum or difference of $n$'th roots, we exploit a formula that comes up in precalculus (usually in the section on synthetic division, although some books give this after the formulas for the difference of squares and the difference of cubes): $$ a^{n} - b^{n} \;\; = \;\; (a-b)\left[ a^{n-1} \; + \; a^{n-2}b \; + \; \dots \; + \; b^{n-1}\right] $$ We can use this formula to rationalize the numerator of the $h$-difference quotient above by letting $a = (x+h) ^{\frac{m}{n}}$ and $b = x^{\frac{m}{n}}.$

To carry this out, multiply the numerator and denominator by the appropriate conjugate, namely $a^{n-1} + \dots + b^{n-1}$ for $a = (x+h) ^{\frac{m}{n}}$ and $b = x^{\frac{m}{n}}.$ Since $a^{n} = \left( (x+h)^{\frac{m}{n}}\right)^{n} = (x+h)^{m}$ and $b^{n} = \left( x^{\frac{m}{n}} \right)^{n} = x^{m},$ we get the following for the $h$-difference quotient of $f:$ $$\frac{a^{n} - b^{n}}{h \cdot (conjugate \;\; terms)} \;\; = \;\; \frac{(x+h)^{m} - x^{m}}{h \cdot (conjugate \;\; terms)} $$ By the way, there is no need to actually write out the "conjugate terms" in full. Like a lot of theoretical math computations, it often pays to avoid doing knee-jerk calculations and wait until later to see whether you really have to perform the calculations or to what degree of completeness the calculations have to be performed.

In the spirit of trying to avoid unnecessary calculations, note that since we're going to let $h \rightarrow 0,$ we only need to write down the dominant term in the numerator and the dominant term in the denominator.

The numerator is $$ x^{m} \; + \; mx^{m-1}h \; + \; (higher \; order \; terms \; in \; h) \; - \; x^{m} $$ $$ =\;\; mx^{m-1}h \; + \; (terms \; with \; factor \; h^{2})$$ $$ = \;\; h \cdot \left[ mx^{m-1} \; + \; (terms \; with \; factor \; h) \right] $$ It is important to note that there is a fixed number of terms in the group "terms with factor $h."$ In fact, there are exactly $m-1$ such terms. As an example similar to this in which the number of "terms with factor $h"$ is not fixed, consider $\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}.$ In contrast to what we have, note that as $n \rightarrow \infty$ each of the terms in the sum $\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}$ approaches zero, but the sum itself does not approach zero. You can use equal width Riemann sums and the definition of the definite integral to determine what the limit is.

From the work we've done, it follows that the $h$-difference quotient is $$ \frac{h \cdot \left[ mx^{m-1} \; + \; (terms \; with \; factor \; h) \right]}{h \cdot (conjugate \; terms)}$$ Multiplicatively cancel the factors of $h$ and let $h \rightarrow 0.$ (In our case, just put $h = 0.)$ The result will be $\frac{d}{dx}\left(x^{\frac{m}{n}}\right)$: $$\frac{d}{dx}\left(x^{\frac{m}{n}}\right) \;\; = \;\; \frac{m \cdot x^{m-1}}{conjugate \; terms \; when \; h=0} $$ The "conjugate terms when $h=0$" consist of $n$ identical terms of $x^{\frac{m}{n}},$ each raised to the $n-1$ power, or

$$ (conjugate \; terms \; when \; h=0) \;\; = \;\; n \cdot \left(x^{\frac{m}{n}}\right)^{n-1} \; = \; n \cdot x^{\frac{m(n-1)}{n}} $$ Hence, the derivative is $$ \frac{m \cdot x^{m-1}}{conjugate \; terms \; when \; h=0} \;\; = \;\; \frac{m \cdot x^{m-1}}{n \cdot x^{\frac{m(n-1)}{n}}} $$ $$ = \;\; \frac{m}{n} \cdot \frac{x^{m-1}}{x^{\frac{m(n-1)}{n}}}$$ $$= \;\; \frac{m}{n} \cdot x^{(m-1) - \frac{m(n-1)}{n}} $$ $$ = \;\; \frac{m}{n} \cdot x^{\frac{(m-1)n - m(n-1)}{n}} $$ $$ = \;\; \frac{m}{n} \cdot x^{\frac{mn - n - mn + m}{n}} $$ $$ = \;\; \frac{m}{n} \cdot x^{\frac{m – n}{n}} $$ $$ = \frac{m}{n} \cdot x^{\frac{m}{n} - \frac{n}{n}} $$ $$ = \;\; \frac{m}{n} \cdot x^{\frac{m}{n} – 1} $$ This gives the differentiation rule for positive rational powers. For negative rational powers, things are a little messier, but nothing creative is needed.

Assume $m$ and $n$ are positive integers. Then $$ \frac{(x+h)^{-\frac{m}{n}} \; - \; x^{-\frac{m}{n}}}{h} $$ $$ = \;\; \frac{(x+h)^{-\frac{m}{n}} \; - \; x^{-\frac{m}{n}}}{h} \; \cdot \; \frac{x^{\frac{m}{n}}(x+h)^{\frac{m}{n}}}{x^{\frac{m}{n}}(x+h)^{\frac{m}{n}}} $$ $$ = \;\; \frac{x^{\frac{m}{n}} \; - \; (x+h)^{\frac{m}{n}}}{h \cdot x^{\frac{m}{n}} \cdot (x+h)^{\frac{m}{n}}} $$ Now continue in the same manner as above. The extra two factors in the denominator just come along for the ride. At the end, these extra two factors contribute to the exponent calculations in such a way that, when you're done, you should find that the derivative of $x^{-\frac{m}{n}}$ is $-\frac{m}{n}x^{-\frac{m}{n} - 1}.$

  • $\begingroup$ Thanks for your answer, you clearly put a lot of effort in! $\endgroup$ – Samuel Tan May 16 '13 at 9:10

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