Question

Find the derivative of the cube
root of x using first principle.
​

Answer 1

We have to compute the limit

`\displaystyle \lim_(h\to 0) \frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}`

We want to use the cube difference formula:

`a^3-b^3 = (a - b)(a^2 + ab + b^2)`

So, we can multiply both numerator and denominator as follows

`\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h} = \frac{\left(\sqrt[3]{x+h}-\sqrt[3]{x}\right)\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)}{h\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)}`

The numerator is now the difference of cubes, so we have

`\frac{x+h-x}{h\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)} = \frac{h}{h\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)} = \frac{1}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}`

As h approaches zero, we have

`\displaystyle \lim_(h\to 0)\frac{1}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}} = \frac{1}{\sqrt[3]{(x+0)^2}+\sqrt[3]{x(x+0)}+\sqrt[3]{x^2}} = (1)/(3√(x^2))`