JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(y = {e^{nx}}\), then \(\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)\left( {\frac{{{d^2}x}}{{d{y^2}}}} \right)\) is equal to
- A \(ne^{nx}\)
- B \(ne^{-nx}\)
- C \(1\)
- D \(-ne^{-nx}\)
Answer & Solution
Correct Answer
(D) \(-ne^{-nx}\)
Step-by-step Solution
Detailed explanation
Given that, \(y = {e^{nx}}\) Differeniating both sides with respect to \(x\) \(\frac{{dy}}{{dx}} = n{e^{nx}}\) Again differentiating w.r.t \('x'\), \(\frac{{{d^2}y}}{{d{x^2}}} = n.n{e^{nx}} = {n^2}{e^{nx}}\,\,\,\,\,\,\,\,....\left( 1 \right)\) \(y = {e^{nx}}\)…
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