A) p'(x)p"'(x)
B) p"(x)p'"(x)
C) p(x)p"'(x)
D) A constant
Correct Answer: C
Solution :
| [c] Given that \[{{y}^{2}}=p(x)\]Differentiating |
| \[\Rightarrow 2y{{y}_{1}}=p'(x)\]\[\left[ here{{y}_{1}}=\frac{dy}{dx} \right]\] |
| \[\Rightarrow 2{{y}_{1}}=\frac{p'(x)}{y}\] |
| Differentiating again, |
| \[\Rightarrow 2{{y}_{2}}=\frac{yp''(x)-p'(x){{y}_{1}}}{{{y}^{2}}},\left[ {{y}_{2}}=\frac{{{d}^{2}}y}{d{{x}^{2}}} \right]\] |
| \[\Rightarrow 2{{y}_{2}}=\frac{yp''(x)-\frac{p'(x).p'(x)}{2y}}{{{y}^{2}}}\] |
| \[=\frac{2{{y}^{2}}p''(x)-p'(x){{)}^{2}}}{2{{y}^{3}}}\] |
| \[\Rightarrow 2{{y}^{3}}{{y}_{2}}=\frac{1}{2}[2{{y}^{2}}p''(x)-{{(p'(x))}^{2}}]\] |
| \[\Rightarrow 2{{y}^{3}}{{y}_{2}}=\frac{1}{2}[2p(x)p''(x)-{{(p'(x))}^{2}}]\] |
| \[\Rightarrow \,\,\,2\frac{d}{dx}({{y}^{3}}{{y}_{2}})\] |
| \[=\frac{1}{2}[2p'(x)p''(x)+2p(x)p'''(x)-2p'(x)p''(x)]\] |
| \[=p(x)p'''(x)\] |
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