Answer

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Hint: In this question, we have to find derivatives of each term of the summation and apply the relevant derivative and basic algebra formulae.

Let the function \[g(x)={{x}^{n}}+a{{x}^{n-1}}+{{a}^{2}}{{x}^{n-2}}+.........+{{a}^{n-1}}x+{{a}^{n}}\]

The derivative of \[g(x)=\dfrac{dg(x)}{dx}={{g}^{1}}(x)\]

\[\begin{align}

& \Rightarrow {{g}^{1}}(x)=\dfrac{d\left( {{x}^{n}}+a{{x}^{n-1}}+{{a}^{2}}{{x}^{n-2}}+.........+{{a}^{n-1}}x+{{a}^{n}} \right)}{dx} \\

& \\

\end{align}\]

We know that derivative of the summation of terms is equal to the summation of the derivation of terms.

Thus, the above function can be written as

\[{{g}^{1}}(x)=\dfrac{d{{x}^{n}}}{dx}+\dfrac{d\left( a{{x}^{n-1}} \right)}{dx}+\dfrac{d\left( {{a}^{2}}{{x}^{n-2}} \right)}{dx}+........\dfrac{d\left( {{a}^{n-1}}x \right)}{dx}+\dfrac{d\left( {{a}^{n}} \right)}{dx}\]

We know that derivative of \[{{x}^{n}}\]is given by the formula \[\dfrac{d({{x}^{n}})}{dx}=n.{{x}^{n-1}}\] and the derivative of constant is given by the formula \[\dfrac{da}{dx}=0\].

The formula for derivative of the product of two functions is given by \[\dfrac{d\left( uv \right)}{dx}=u\dfrac{d(v)}{dx}+v\dfrac{d(u)}{dx}\]

Where \[u,v\]can be functions of \[x\] or constant values.

Thus, we get \[\dfrac{d\left( a{{x}^{n-1}} \right)}{dx}={{x}^{n-1}}\dfrac{d\left( a \right)}{dx}+a\dfrac{d({{x}^{n-1}})}{dx}\]

\[={{x}^{n-1}}(0)+a(n-1)({{x}^{(n-1)-1}})\]

\[=a(n-1)({{x}^{n-2}})\]

By applying the formula for other terms of the equation we get,

\[{{g}^{1}}(x)=n.{{x}^{n-1}}+a(n-1){{x}^{n-2}}+{{a}^{2}}(n-2){{x}^{n-3}}.........+{{a}^{n-1}}(1){{x}^{0}}+0\] ---(1)

We know the formula \[{{x}^{0}}=1\]

\[\therefore {{g}^{1}}(x)=n.{{x}^{n-1}}+a(n-1){{x}^{n-2}}+{{a}^{2}}(n-2){{x}^{n-3}}.........+{{a}^{n-1}}\]

Hence, the derivative of \[{{x}^{n}}+a{{x}^{n-1}}+{{a}^{2}}{{x}^{n-2}}+...+{{a}^{n-1}}x+{{a}^{n}}\]\[=n.{{x}^{n-1}}+a(n-1){{x}^{n-2}}+{{a}^{2}}(n-2){{x}^{n-3}}+...+{{a}^{n-1}}\]

Note: We might mistake the question to be a binomial expansion expression, but the question does not involve the factorial terms of a binomial expansion. The question is an algebraic expression. In the question, since \[a\] is a constant only \[x\] is variable, so the expression should be differentiated with respect to \[x\]. To solve the question, we have to apply the derivative formula of the product of two functions to eliminate the constant values in the derivation, this will ease the procedure of solving. We have to apply basic algebra formulae which are needed in solving the step to arrive at the solution. While solving the question remember to solve it each term wise to avoid overcrowding of the variables.

Let the function \[g(x)={{x}^{n}}+a{{x}^{n-1}}+{{a}^{2}}{{x}^{n-2}}+.........+{{a}^{n-1}}x+{{a}^{n}}\]

The derivative of \[g(x)=\dfrac{dg(x)}{dx}={{g}^{1}}(x)\]

\[\begin{align}

& \Rightarrow {{g}^{1}}(x)=\dfrac{d\left( {{x}^{n}}+a{{x}^{n-1}}+{{a}^{2}}{{x}^{n-2}}+.........+{{a}^{n-1}}x+{{a}^{n}} \right)}{dx} \\

& \\

\end{align}\]

We know that derivative of the summation of terms is equal to the summation of the derivation of terms.

Thus, the above function can be written as

\[{{g}^{1}}(x)=\dfrac{d{{x}^{n}}}{dx}+\dfrac{d\left( a{{x}^{n-1}} \right)}{dx}+\dfrac{d\left( {{a}^{2}}{{x}^{n-2}} \right)}{dx}+........\dfrac{d\left( {{a}^{n-1}}x \right)}{dx}+\dfrac{d\left( {{a}^{n}} \right)}{dx}\]

We know that derivative of \[{{x}^{n}}\]is given by the formula \[\dfrac{d({{x}^{n}})}{dx}=n.{{x}^{n-1}}\] and the derivative of constant is given by the formula \[\dfrac{da}{dx}=0\].

The formula for derivative of the product of two functions is given by \[\dfrac{d\left( uv \right)}{dx}=u\dfrac{d(v)}{dx}+v\dfrac{d(u)}{dx}\]

Where \[u,v\]can be functions of \[x\] or constant values.

Thus, we get \[\dfrac{d\left( a{{x}^{n-1}} \right)}{dx}={{x}^{n-1}}\dfrac{d\left( a \right)}{dx}+a\dfrac{d({{x}^{n-1}})}{dx}\]

\[={{x}^{n-1}}(0)+a(n-1)({{x}^{(n-1)-1}})\]

\[=a(n-1)({{x}^{n-2}})\]

By applying the formula for other terms of the equation we get,

\[{{g}^{1}}(x)=n.{{x}^{n-1}}+a(n-1){{x}^{n-2}}+{{a}^{2}}(n-2){{x}^{n-3}}.........+{{a}^{n-1}}(1){{x}^{0}}+0\] ---(1)

We know the formula \[{{x}^{0}}=1\]

\[\therefore {{g}^{1}}(x)=n.{{x}^{n-1}}+a(n-1){{x}^{n-2}}+{{a}^{2}}(n-2){{x}^{n-3}}.........+{{a}^{n-1}}\]

Hence, the derivative of \[{{x}^{n}}+a{{x}^{n-1}}+{{a}^{2}}{{x}^{n-2}}+...+{{a}^{n-1}}x+{{a}^{n}}\]\[=n.{{x}^{n-1}}+a(n-1){{x}^{n-2}}+{{a}^{2}}(n-2){{x}^{n-3}}+...+{{a}^{n-1}}\]

Note: We might mistake the question to be a binomial expansion expression, but the question does not involve the factorial terms of a binomial expansion. The question is an algebraic expression. In the question, since \[a\] is a constant only \[x\] is variable, so the expression should be differentiated with respect to \[x\]. To solve the question, we have to apply the derivative formula of the product of two functions to eliminate the constant values in the derivation, this will ease the procedure of solving. We have to apply basic algebra formulae which are needed in solving the step to arrive at the solution. While solving the question remember to solve it each term wise to avoid overcrowding of the variables.

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