Question

Find the \[{n^{th}}\] derivative of \[{\left( {x + 1} \right)^n}\]
A. \[\left( {n - 1} \right)!\]
B. \[\left( {n + 1} \right)!\]
C. \[n!\]
D. \[n{\left[ {\left( {n - 1} \right)} \right]^{n - 1}}\]

Answer 1

Hint: For finding the \[{n^{th}}\] derivative of \[{\left( {x + 1} \right)^n}\]. We will be differentiating \[{\left( {x + 1} \right)^n}\] for n times which means we will be differentiating it for the first derivative , second derivative , third derivative and so on.

Formula Used: We will be using basic formula of Differentiation which is \[\dfrac{d}{dx}{x^n} = n{\left( x \right)^{n - 1}}\]

Complete step by step solution
For finding the first derivative of \[{\left( {x + 1} \right)^n}\]we will differentiate it with respect to \[x\] .
Let \[y = {\left( {x + 1} \right)^n}\]
\[\therefore \dfrac{{dy}}{{dx}} = n{\left( {x + 1} \right)^{n - 1}}\]
Now we will be finding the second derivative of \[{\left( {x + 1} \right)^n}\]:-
\[\therefore \dfrac{{{d^2}y}}{{d{x^2}}} = n\left( {n - 1} \right){\left( {x + 1} \right)^{n - 2}}\]
Similarly we will find the \[{n^{th}}\] derivative of \[{\left( {x + 1} \right)^n}\].
\[\therefore \dfrac{{{d^n}y}}{{d{x^n}}} = n\left( {n - 1} \right)\left( {n - 2} \right)......3.2.1 \times {\left( {x + 1} \right)^{n - n}}\]
\[ = n!\]
Hence the answer is (C) which is\[ n!\].

Note: While solving the question we should stay focused on differentiating \[{\left( {x + 1} \right)^n}\] at least three times and then finding the \[{n^{th}}\] derivative and student should also remember the expansion of \[n!\] to find out which expansion has been created at the \[{n^{th}}\] derivative.