Question

If we have an expression as $y=a{{x}^{n+1}}+b{{x}^{-n}}$ , then find the value $\dfrac{{{x}^{2}}{{d}^{2}}y}{d{{x}^{2}}}=$
$A)n(n-1)y$
$B)n(n+1)y$
$C)ny$
$D){{n}^{2}}y$

Answer 1

Hint: To solve this question we need to have the knowledge of differentiation. In the question given above the function is in terms of the sum of two algebraic functions $a{{x}^{n+1}}$ and $b{{x}^{-n}}$. So to solve the problem we will use the formula of differentiation for $u$ and $v$ which implies $\dfrac{d\left( u+v \right)}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$ . On solving this we get the first differentiation while we will differentiate again to find the second order differentiation of the result we got. The formula used here to solve the question is $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$.

Complete step-by-step solution:
The question ask us to find the double differentiation of the function given to us which is $y=a{{x}^{n+1}}+b{{x}^{-n}}$. Now we will solve the problem by considering this function to be the product of the two other function $u$ and $v$ where $u=a{{x}^{n+1}}$ and $v=b{{x}^{-n}}$. To find the differentiation we will use the formula of the sum of the two function which is $\dfrac{d\left( u+v \right)}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$. On applying the same we get:
\[\Rightarrow \dfrac{d\left( a{{x}^{n+1}}+b{{x}^{-n}} \right)}{dx}=\dfrac{d\left( a{{x}^{n+1}} \right)}{dx}+\dfrac{d\left( b{{x}^{-n}} \right)}{dx}\]
The formula used here to solve the above expression is $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$. On applying the same in the question we get:
\[\Rightarrow \dfrac{d\left( a{{x}^{n+1}}+b{{x}^{-n}} \right)}{dx}=a\left( n+1 \right){{x}^{n+1-1}}+-nb{{x}^{-n-1}}\]
\[\Rightarrow \dfrac{d\left( a{{x}^{n+1}}+b{{x}^{-n}} \right)}{dx}=a\left( n+1 \right){{x}^{n}}-nb{{x}^{-n-1}}\]
\[\Rightarrow \dfrac{dy}{dx}=a\left( n+1 \right){{x}^{n}}-nb{{x}^{-n-1}}\]
The above is the result of one order differentiation, we will again differentiate it to find the double differentiation of the function given in the question. On again differentiating it we get:
$\Rightarrow \dfrac{d\left( \dfrac{dy}{dx} \right)}{dx}=\dfrac{d\left( a\left( n+1 \right){{x}^{n}}-nb{{x}^{-n-1}} \right)}{dx}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( a\left( n+1 \right){{x}^{n}} \right)-d\left( nb{{x}^{-n-1}} \right)}{dx}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( a\left( n+1 \right){{x}^{n}} \right)}{dx}-\dfrac{d\left( nb{{x}^{-n-1}} \right)}{dx}$
On applying the same formula as above in the expression we get:
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( a\left( n+1 \right){{x}^{n}} \right)}{dx}-\dfrac{d\left( nb{{x}^{-n-1}} \right)}{dx}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=an\left( n+1 \right){{x}^{n-1}}-n\left( -n-1 \right)b{{x}^{-n-1-1}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=an\left( n+1 \right){{x}^{n-1}}+n\left( n+1 \right)b{{x}^{-n-2}}$
Multiplying both side of the expression with ${{x}^{2}}$ to find the value of the expression given in the question, we get:
$\Rightarrow {{x}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{x}^{2}}\left( an\left( n+1 \right){{x}^{n-1}} \right)+{{x}^{2}}\left( n\left( n+1 \right)b{{x}^{-n-2}} \right)$
On multiplying we get:
$\Rightarrow {{x}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=an\left( n+1 \right){{x}^{n-1+2}}+n\left( n+1 \right)b{{x}^{-n-2+2}}$
$\Rightarrow {{x}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=an\left( n+1 \right){{x}^{n+1}}+n\left( n+1 \right)b{{x}^{-n}}$
$\Rightarrow {{x}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=n\left( n+1 \right)\left( a{{x}^{n+1}}+b{{x}^{-n}} \right)$
On substituting the above function with $y$ we get:
$\Rightarrow {{x}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=n\left( n+1 \right)y$
$\therefore $ If $y=a{{x}^{n+1}}+b{{x}^{-n}}$ , then $\dfrac{{{x}^{2}}{{d}^{2}}y}{d{{x}^{2}}}$ is $B)n(n+1)y$.

Note: We can check whether the solution we got is correct or not. This could be done by dividing the answer with ${{x}^{2}}$ and then integrating the result twice. If the answer comes the same as that given in the question it implies the answer is correct.