KCET · Maths · Application of Derivatives
If \(y=2 x^{n+1}+\frac{3}{x^{n}}\), then \(x^{2} \frac{d^{2 y}}{d x^{2}}\) is
- A \(6 n(n+1) y\)
- B \(n(n+1) y\)
- C \(x \frac{d y}{d x}+y\)
- D \(y\)
Answer & Solution
Correct Answer
(B) \(n(n+1) y\)
Step-by-step Solution
Detailed explanation
We have, \(y\)
\(\begin{aligned}
y &=2 x^{n+1}+\frac{3}{x^{n}} \\
&=2 x^{n+1}+3 x^{-n}...(i)
\end{aligned}\)
On differentiating Eq. (i) both sides w.r.t. \(x\),
we get
\(\frac{d y}{d x}=2(n+1) x^{n}+3(-n) x^{-n-1}\)
Again differentiate w.r.t. to \(x\), then we get
\(\frac{d^{2} y}{d x^{2}}=2(n+1)(n) x^{n-1}+3(-n)(-n-1) x^{-n-2}\)
\(=n(n+1)\left(2 x^{n-1}+3 x^{-n-2}\right)\)
\(\Rightarrow \quad x^{2} \frac{d^{2} y}{d x^{2}}=n(n+1)\left(2 x^{n+1}+\frac{3}{x^{n}}\right)\)
\(\Rightarrow \quad x^{2} \frac{d y}{d x}=n(n+1) y\) [using Eq. (i)]
\(\begin{aligned}
y &=2 x^{n+1}+\frac{3}{x^{n}} \\
&=2 x^{n+1}+3 x^{-n}...(i)
\end{aligned}\)
On differentiating Eq. (i) both sides w.r.t. \(x\),
we get
\(\frac{d y}{d x}=2(n+1) x^{n}+3(-n) x^{-n-1}\)
Again differentiate w.r.t. to \(x\), then we get
\(\frac{d^{2} y}{d x^{2}}=2(n+1)(n) x^{n-1}+3(-n)(-n-1) x^{-n-2}\)
\(=n(n+1)\left(2 x^{n-1}+3 x^{-n-2}\right)\)
\(\Rightarrow \quad x^{2} \frac{d^{2} y}{d x^{2}}=n(n+1)\left(2 x^{n+1}+\frac{3}{x^{n}}\right)\)
\(\Rightarrow \quad x^{2} \frac{d y}{d x}=n(n+1) y\) [using Eq. (i)]
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