Obtaining a derivative using limit definition We have the following limit $$ \lim_{h \to 0} \dfrac{(x+h)^{1/4}-x^{1/4}}{h}$$
I want to find this limit (which I figure is just the derivative of $x^{1/4}$) using only elementary methods (algebra, mostly).
So you can rewrite the function as $\dfrac{4}{h}(\sqrt{x+h} - \sqrt{x})$, but you still have that $h$ in the denominator which makes it impossible to take the limit. I can't seem to rewrite this in a way that I get the expected answer, can anyone give a hand?
 A: Use substitution: set $y=x^{\tfrac14}, \enspace k=(x+h)^{\tfrac14}-x^{\tfrac14}$. Note $k\to 0\;$ as $h\to 0$. Let's rewrite the variation rate:
$$\frac{(x+h)^{1/4}-x^{1/4}}{h}=\frac k{(y+k)^4-y^4}=\frac k{4y^3k+6y^2k^2+4yk^3}=\frac 1{4y^3+6y^2k+4yk^2},$$
which tends to
$$\frac1{4y^3}=\frac1{4x^{3/4}}$$
as $h$ (or $k$) tends to $0$.
A: Hint:
Use the fact that
$$\frac{(x+h)^{1/4}-x^{1/4}}{h}=\frac{\sqrt[4]{x+h}-\sqrt[4]{x}}{h}$$
Multiply both terms in the fraction by $$(\sqrt[4]{x+h})^3+(\sqrt[4]{x+h})^2\sqrt[4]{x}+(\sqrt[4]{x+h})(\sqrt[4]{x})^2+(\sqrt[4]{x})^3$$
And simplify:
$$\lim_{h\to 0}\frac{(x+h)^{1/4}-x^{1/4}}{h}=\lim_{h\to 0}\frac{(\sqrt[4]{x+h})^4-(\sqrt[4]{x})^4}{h\left[(\sqrt[4]{x+h})^3+(\sqrt[4]{x+h})^2\sqrt[4]{x}+(\sqrt[4]{x+h})(\sqrt[4]{x})^2+(\sqrt[4]{x})^3\right]}$$
A: Using Binomial Expansion,
 $\frac{(x+h)^{1/4}-x^{1/4}}{h} = \frac{x^{1/4}+(1/4)x^{(1/4-1)}+O(h^2)-x^{1/4}}{h}=\frac{(1/4)x^{(1/4-1)}+O(h^2)}{h} = (1/4)x^{(1/4-1)}+O(h)$
$lim_{h\rightarrow0} (1/4)x^{(1/4-1)}+lim_{h\rightarrow0}O(h) = (1/4)x^{(1/4-1)} = (1/4)x^{-5/4} $
A: $$\lim_{h \to 0} \dfrac{(x+h)^{1/4}-x^{1/4}}{h}$$
Let $y = x + \frac12h$.
$$\lim_{h \to 0} \dfrac{(y+\frac12h)^{1/4}-(y-\frac12h)^{1/4}}{h}$$
Then,
$$
\lim_{h \to 0} \dfrac{(y+\frac12h)^{1/4}-(y-\frac12h)^{1/4}}{h}
= \lim_{h \to 0}\left[\dfrac{(y+\frac12h)-(y-\frac12h)}{h} \div \left((y+\frac12h)^{1/2}+(y-\frac12h)^{1/2}\right)\left((y+\frac12h)^{1/4}+(y-\frac12h)^{1/4}\right)\right]
= \lim_{h \to 0}\left[1 \div \left((y+\frac12h)^{1/2}+(y-\frac12h)^{1/2}\right)\left((y+\frac12h)^{1/4}+(y-\frac12h)^{1/4}\right)\right]
$$
Simply plug in $0$ for $h$ and unsubstitute and you will get $\frac14x^{-\frac34}$

Note: I am aware that the substitution was arbitrary and does not really add anything. I like it like that.
A: Successively multiplying top and bottom by conjugates, we have $$\frac{(x+h)^{\frac 14}-x^{\frac 14}}{h}=\frac{(x+h)^{\frac 12}-x^{\frac 12}}{h((x+h)^{\frac 14}+x^{\frac 14})}$$
$$=\frac{(x+h)-x}{h((x+h)^{\frac 14}+x^{\frac 14})((x+h)^{\frac 12}+x^{\frac 12})}$$
Now cancel $h$ from top and bottom and let $h\rightarrow 0$, and we have $$\frac{1}{(2x^{\frac 14})(2x^{\frac 12})}=\frac 14 x^{-\frac 34}$$
