TS EAMCET · Maths · Differentiation
If \(y=\tan ^{-1}(\sin \sqrt{x})+\operatorname{cosec}^{-1}\left(e^{2 x+1}\right)\), then \(\frac{d y}{d x}=\)
- A \(\frac{1}{\sqrt{x}\left(1+\sin ^2 \sqrt{x}\right)}+\frac{1}{\sqrt{e^{4 x+2}+1}}\)
- B \(\frac{\cos \sqrt{x}}{2 \sqrt{x}\left(1+\sin ^2 \sqrt{x}\right)}-\frac{2}{\sqrt{e^{4 x+2}-1}}\)
- C \(\frac{\cos \sqrt{x}}{\left(1+\sin ^2 \sqrt{x}\right)}+\frac{2}{\sqrt{e^{4 x+2}+1}}\)
- D \(\frac{1}{2 \sqrt{x}} \frac{\cos \sqrt{x}}{\left(1+\sin ^2 \sqrt{x}\right)}-\frac{1}{\sqrt{e^{2 x+1}-1}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\cos \sqrt{x}}{2 \sqrt{x}\left(1+\sin ^2 \sqrt{x}\right)}-\frac{2}{\sqrt{e^{4 x+2}-1}}\)
Step-by-step Solution
Detailed explanation
\(y=\tan ^{-1}(\sin \sqrt{x})+\operatorname{cosec}^{-1}\left(e^{2 x+1}\right)\) So, \(\frac{d y}{d x}=\frac{1}{1+\sin ^2 \sqrt{x}} \cdot \frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{1 \cdot e^{2 x+1} \cdot 2}{e^{2 x+1} \sqrt{e^{4 x+2}-1}}\)…
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