MHT CET · Maths · Differentiation
If \(y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)\), then \(\frac{d y}{d x}=\)
- A \(\frac{3}{9 x^2+12 x+5}+\frac{1}{2 x^2-2 x+1}\)
- B \(\frac{1}{9 x^2+12 x+5}+\frac{1}{4 x^2-4 x+2}\)
- C \(\frac{1}{9 x^2+12 x+5}-\frac{1}{4 x^2-4 x+2}\)
- D \(\frac{3}{9 x^2+12 x+5}-\frac{1}{2 x^2-2 x+1}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{9 x^2+12 x+5}+\frac{1}{2 x^2-2 x+1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)=\tan ^{-1} \frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)} \\ & \Rightarrow y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+(3 x+2)} \times 3+\frac{1}{1+(2 x-1)^2} \times 2 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{9 x^2+12 x+5}+\frac{2}{4 x^2-4 x+2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3}{9 x^2+12 x+5}+\frac{1}{2 x^2-2 x+1}\end{aligned}\)
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