Question

Differentiate from first principle:

$\left(iii\right)\frac{1}{x^3}$

Answer 1

Given: $f\left(x\right)=\frac{1}{x^3}$

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Putting $f\left(x\right)$ the above expression, we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{1}{(x+h{)}^3}-\frac{1}{x^3}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x^3-(x+h{)}^3}{h(x+h{)}^3{x}^3}$

Applying formula,

$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3,$ we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x^3-x^3-3x^{2}h-3x{h}^2-h^3}{h(x+h{)}^3{x}^3}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{-3x^{2}h-3x{h}^2-h^3}{h(x+h{)}^3{x}^3}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{-3x^2-3xh-h^2}{(x+h{)}^3{x}^3}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-3x^2}{x^6}$

$\Rightarrow f^{\prime }\left(x\right)=-3x^{-4}$

Therefore, the derivative of $\frac{1}{x^3}is-3x^{-4}.$