<s>[INST] how do you differentiate x^(3/4) using first principle $$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac{3}{4}}-(x)^{\frac{3}{4}}\Bigr)}{h}$$
I understand the process till 
$$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac34}-(x)^{\frac{3}{4}}\Bigr)}{h} * \frac{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}$$
and post expansion
$$\lim_{h\to 0}\frac{\Bigl(h^3+3h^2x+3x^2h\Bigr)}{{h}\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)\Bigl((x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}\Bigr)}$$
but beyond this i am unable to reduce to:
$\frac 34\cdot x^{\frac{-1}{4}}$
 [/INST] Why don't you expand from the start
$$\lim_{h\to0}\dfrac{(x+h)^n-x^n}h=x^n\cdot\lim_{h\to0}\dfrac{\left(1+\dfrac hx\right)^n-1}h$$
Now use Binomial series
Alternatively, set $$(x+h)^{1/4}=a,x+h=a^4; x^{1/4}=b, x=b^4$$
$$\lim_{h\to0}\dfrac{(x+h)^{3/4}-x^{1/4}}h=\lim_{a\to b}\dfrac{a^3-b^3}{a^4-b^4}=\lim_{a\to b}\dfrac{a^2+ab+b^2}{a^3+a^2b+ab^2+b^3}=\dfrac{3b^2}{4b^3}=\dfrac3{4b}=\dfrac3{4x^{1/4}}$$
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