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Question: The \({{n}^{th}}\) derivative of \({{\left( x+1 \right)}^{n}}\) is equal to 1\. \(\left( n-1 \righ...

The nth{{n}^{th}} derivative of (x+1)n{{\left( x+1 \right)}^{n}} is equal to
1. (n−1)!\left( n-1 \right)!
2. (n+1)!\left( n+1 \right)!
3. n!n!
4. n[(n+1)]n−1n{{\left[ \left( n+1 \right) \right]}^{n-1}}

Explanation

Solution

To find the nth{{n}^{th}} derivative of the given function we will differentiate the given function with respect to x and find derivatives up to the order of 3. Then by analyzing the pattern after combining the terms obtained we will get the desired answer. We will use the following formula to differentiate the given function
ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}

Complete step-by-step solution:
We have been given a function (x+1)n{{\left( x+1 \right)}^{n}}.
We have to find the nth{{n}^{th}} derivative of the given function.
Let us assume that the given function is
⇒y=(x+1)n\Rightarrow y={{\left( x+1 \right)}^{n}}
Now, let us differentiate the given function with respect to x, then we will get
⇒dydx=ddx(x+1)n\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}{{\left( x+1 \right)}^{n}}
Now, we know that the power formula of differentiation is given by
ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}
Now, applying the formula to the given function we will get
⇒dydx=n(x+1)n−1\Rightarrow \dfrac{dy}{dx}=n{{\left( x+1 \right)}^{n-1}}
Now, again differentiating the above obtained derivative with respect to x we will get
⇒y′′=n(n−1)(x+1)n−2\Rightarrow y''=n\left( n-1 \right){{\left( x+1 \right)}^{n-2}}
Now, again differentiating the above obtained derivative with respect to x we will get
⇒y′′′=n(n−1)(n−2)(x+1)n−3\Rightarrow y'''=n\left( n-1 \right)\left( n-2 \right){{\left( x+1 \right)}^{n-3}}
Now, the nth{{n}^{th}} derivative of the given function will be
⇒yn=n(n−1)(n−2)......(x+1)n−n\Rightarrow {{y}^{n}}=n\left( n-1 \right)\left( n-2 \right)......{{\left( x+1 \right)}^{n-n}}
Therefore we can write the above series as
⇒yn=n(n−1)(n−2)......(x+1)0 ⇒yn=n! \begin{aligned} & \Rightarrow {{y}^{n}}=n\left( n-1 \right)\left( n-2 \right)......{{\left( x+1 \right)}^{0}} \\\ & \Rightarrow {{y}^{n}}=n! \\\ \end{aligned}
Hence the nth{{n}^{th}} derivative of the given function is n!n!.
Option 3 is the correct answer.

Note: By generalizing the pattern and combining the terms obtained we reach the conclusion that the obtained pattern is of factorial. If options are not given in the question we can end up the solution simplify by finding the nth{{n}^{th}} derivative of the given function as yn=n(n−1)(n−2)......(x+1)n−n{{y}^{n}}=n\left( n-1 \right)\left( n-2 \right)......{{\left( x+1 \right)}^{n-n}}.