WBJEE · Maths · Differentiation
Let \(\mathrm{y}=\frac{\mathrm{x}^{2}}{(\mathrm{x}+1)^{2}(\mathrm{x}+2)}\). Then \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) is
- A \(2\left[\frac{3}{(x+1)^{4}}-\frac{3}{(x+1)^{3}}+\frac{4}{(x+2)^{3}}\right]\)
- B \(3\left[\frac{2}{(x+1)^{3}}+\frac{4}{(x+1)^{2}}-\frac{5}{(x+2)^{3}}\right]\)
- C \(\frac{6}{(x+1)^{3}}-\frac{4}{(x+1)^{2}}+\frac{3}{(x+1)^{3}}\)
- D \(\frac{7}{(x+1)^{3}}-\frac{3}{(x+1)^{2}}+\frac{2}{(x+1)^{3}}\)
Answer & Solution
Correct Answer
(A) \(2\left[\frac{3}{(x+1)^{4}}-\frac{3}{(x+1)^{3}}+\frac{4}{(x+2)^{3}}\right]\)
Step-by-step Solution
Detailed explanation
Hint: By partial fraction technique \(\begin{array}{l} y=\frac{x^{2}}{(x+1)^{2}(x+2)}=\frac{4}{(x+2)}-\frac{3}{(x+1)}+\frac{1}{(x+1)^{2}} \\ \Rightarrow y^{\prime \prime}=\frac{6}{(x+1)^{4}}-\frac{6}{(x+1)^{3}}+\frac{8}{(x+2)^{3}} \end{array}\)
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