MHT CET · Maths · Differentiation
If \(y=\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)+\tan ^{-1}\left(\frac{3+8 x}{3-3 x}\right)\), then \(\frac{d y}{d x}=\)
- A \(\frac{1}{1+25 x^2}\)
- B \(\frac{5}{1+25 x^2}\)
- C \(\frac{1}{1+5 x^2}\)
- D \(\frac{5}{1+5 x^2}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{1+25 x^2}\)
Step-by-step Solution
Detailed explanation
\(y=\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)+\tan ^{-1}\left(\frac{3+8 x}{8-3 x}\right)=\) \(\tan ^{-1}\left(\frac{5 x-x}{1+5 x \cdot x}\right)+\tan ^{-1}\left(\frac{\frac{3}{8}+x}{1-\frac{3}{8} \cdot x}\right) \)
\( =\tan ^{-1}(5 x)-\tan ^{-1}(x)+\tan ^{-1}\left(\frac{3}{8}\right)+\tan ^{-1}(x) \)
\( \Rightarrow y=\tan ^{-1}(5 x)+\tan ^{-1}\left(\frac{3}{8}\right) \)
\( \Rightarrow \frac{d y}{d x}=\frac{1}{1+(5 x)^2} \times 5+0=\frac{5}{1+25 x^2}\)
\()
\( =\tan ^{-1}(5 x)-\tan ^{-1}(x)+\tan ^{-1}\left(\frac{3}{8}\right)+\tan ^{-1}(x) \)
\( \Rightarrow y=\tan ^{-1}(5 x)+\tan ^{-1}\left(\frac{3}{8}\right) \)
\( \Rightarrow \frac{d y}{d x}=\frac{1}{1+(5 x)^2} \times 5+0=\frac{5}{1+25 x^2}\)
\()
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