Question

What is the derivative of ${{x}^{n}}$?

Answer 1

This kind of question are solved based on the concept of differentiation or else by using the limit definition of derivation we can find out the derivative of ${{x}^{n}}$. Assuming ‘h’ as the small change in x and hence find the small change in the function f(x) given as f(x+h). now, the formula for limit definition of derivative is $f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$, substituting the value of the given function and simplifying the limit to get the answer.

Complete step-by-step solution:
Let us solve the given question
Here we have the function $f\left( x \right)={{x}^{n}}$ and we have to find its derivative. Let us use the formula for the limit definition of derivative to get the answer.
The definition of derivative of a function is defined as the rate of change of function. Mathematically we have,
$f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$
We assuming $f\left( x \right)={{x}^{n}}$ and $f\left( x+h \right)={{\left( x+h \right)}^{n}}$
Substituting the above functions in the formula, we can get
$f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{{{\left( x+h \right)}^{n}}-{{x}^{n}}}{h} \right)$
By using the binomial expansion, we can solve the given function
$\Rightarrow \displaystyle \lim_{h \to 0}\left( \dfrac{{{x}^{n}}+nh{{x}^{n-1}}+\dfrac{n\left( n-1 \right)}{2!}{{h}^{2}}{{x}^{n-2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{h}^{3}}{{x}^{n-2}}+.......-{{x}^{n}}}{h} \right)$
In the above, ${{x}^{n}}$ and $-{{x}^{n}}$ are cancelled, then
$\Rightarrow \displaystyle \lim_{h \to 0}\left( \dfrac{nh{{x}^{n-1}}+\dfrac{n\left( n-1 \right)}{2}{{h}^{2}}{{x}^{n-2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}{{h}^{3}}{{x}^{n-2}}+.......}{h} \right)$
Dividing the total function with ‘h’,
$\Rightarrow \displaystyle \lim_{h \to 0}\left( n{{x}^{n-1}}+\dfrac{n\left( n-1 \right)}{2}h{{x}^{n-2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{6}{{h}^{2}}{{x}^{n-2}}+... \right)$
Limit h tends to zero, hence we have to substitute $h=0$.
Then the whole function becomes
$\therefore f'\left( x \right)=n{{x}^{n-1}}.$
The derivative of ${{x}^{n}}$is $n{{x}^{n-1}}$.

Note: Students should note that the limit definition of derivative is also known as the first principle of differentiation. The given function ${{x}^{n}}$ is solved for rationals; we use the chain rule and for irrationals, we use implicit differentiation.
By using the implicit differentiation, we can prove for all real numbers, including the irrationals.
Let, $y={{x}^{n}}$
Applying natural logarithm on both sides
$\ln \left( y \right)=n\cdot \ln \left( x \right)$
It becomes,
$\dfrac{y'}{y}=\dfrac{n}{x}$
$y'=\dfrac{ny}{x}=\dfrac{n{{x}^{n}}}{x}=n{{x}^{n-1}}$.
The derivative of the derivative of ${{x}^{n}}$is $n{{x}^{n-1}}$.