Question

Differentiate from the first principle the cube root of x plus cube root of x.

Answer 1

Given question is to differentiate $x^{1/3}+x^{1/3} =2x^{1/3}$

 Using the first principle

Suppose $f(x)= x^{1/3}$, then we need to evaluate

$lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

Putting the value of f(x) we get,

$lim_{h\to0} \frac{(x+h)^{1/3}-x^{1/3}}{h}$

Multiplying top and bottom by

$(x+h)^{2/3}+ x^{1/3}(x+h)^{1/3}+x^{2/3}$

Now combine the like terms after multiplying numerator and denominator, many of them will cancel out, we get

$lim_{h\to0} \frac{x+h-x}{h((x+h)^{2/3}+(x+h)^{1/3}x^{1/3}+x^{2/3})}$

Now, cancel x's out on the top:

$lim_{h\to0}\frac{h}{h((x+h)^{2/3}+(x+h)^{1/3}x^{1/3}+x^{2/3})}$

$lim_{h\to0} \frac{1}{(x+h)^{2/3}+(x+h)^{1/3}x^{1/3}+x^{2/3}}$

Now, take the limit, we will get

$\frac{1}{x^{2/3}+x^{2/3}+x^{2/3}} =\frac{1}{3x^{2/3}}$

Thus, the derivative of $f(x) = x^{1/3}$ is $f\prime(x)=\frac{1}{3x^{2/3}}$

So, the derivative of $2x^{1/3}$ will be equal to $2f\prime(x)=\frac{2}{3x^{2/3}}$