algebra-precalculuscalculusderivatives
Find the derivative of function $f(x) = \sqrt{x} + \dfrac{1}{x^3}$ from the first principles.
I tried to use the formula
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}.
$$
I try to expand for the $x^3$ first but it looks more and more complicated. The answer end up with $0$.
For concreteness, let's first take $k=3$.
We use the factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$.
Here we have $(x+a+h)^3-(x+a)^3=h((x+a+h)^2+(x+a+h)(x+a)+(x+a)^2)$.
The difference quotient is then
$$\frac{(x+a+h)^3-(x+a)^3}{h}=((x+a+h)^2+(x+a+h)(x+a)+(x+a)^2).$$
As $h$ goes to zero, this becomes $3(x+a)^2$, as desired.
How does this generalize? We have similar factorizations for all $k$:
$$x^k-y^k=(x-y)(x^k+x^{k-1}y+\cdots xy^{k-1}+ y^k)$$
where there are $k$ terms in the second parenthesis. This factorization is easy to check, because the terms all telescope and cancel.
So what I did for $k=3$ can be repeated for general $k$, and when $h$ goes to zero you get $k$ terms of $(x+a)^{k-1}$ added together, which is exactly the same derivative you get using the chain rule.
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 $-\frac{m}{n}x^{-\frac{m}{n} - 1}.$
Best Answer
Hint: (I'm assuming you mean $\require{cancel}f(x) = \sqrt{x} + \frac1{x^3}$.) Here's the "hard" computation for a few similar functions. You can see the pattern: in the end, there is a factor of $h$ in the numerator which will cancel with the $h$ that you are going to use as the denominator in the difference quotient, and all but the first term of the remaining factor will vanish as $h\to 0$.
For $f(x) = \frac1x$: $$\tfrac1{(x+h)}-\tfrac1{x} = \tfrac x{x(x+h)}-\tfrac{x+h}{x(x+h)}=\tfrac{\cancel{x}-(\cancel{x}+h)}{x(x+h)} = \tfrac{-h}{x(x+h)}$$
For $f(x)=\frac1{x^2}$: $$\tfrac1{(x+h)^2}-\tfrac1{x^2} = \tfrac {x^2}{x^2(x+h)^2}-\tfrac{(x+h)^2}{x^2(x+h)^2}=\tfrac{\cancel{x^2}-(\cancel{x^2}+2hx + h^2)}{x^2(x+h)^2}=\tfrac{-2xh - h^2}{x^2(x+h)^2}=\tfrac{-h(2x + h)}{x^2(x+h)^2}$$
For $f(x) = \frac1{x^3}$ (which is what you are using): $$\tfrac1{(x+h)^3}-\tfrac1{x^3} = \tfrac {x^3}{x^3(x+h)^3}-\tfrac{(x+h)^3}{x^3(x+h)^3}=\tfrac{\cancel{x^3}-(\cancel{x^3}+3hx^2 + 3h^2x + h^3)}{x^3(x+h)^3}=\tfrac{-3hx^2 - 3h^2x - h^3}{x^3(x+h)^3}=\tfrac{-h(3x^2 + 2hx + h^2)}{x^3(x+h)^3}$$
I presume you can handle the $\sqrt{x}$ term without difficulty.