JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(x = \sqrt {{2^{\cos e{c^{ - 1}}t}}} \) and \(y = \sqrt {{2^{se{c^{ - 1}}t}}} (\left| t \right|\,\, \ge \,1\,),\) then \(\frac{{dy}}{{dx}}\) is equal to.
- A \(\frac {y}{x}\)
- B \(-\frac {y}{x}\)
- C \(-\frac {x}{y}\)
- D \(\frac {x}{y}\)
Answer & Solution
Correct Answer
(B) \(-\frac {y}{x}\)
Step-by-step Solution
Detailed explanation
Here, \(\frac{{dx}}{{dt}} = \frac{1}{{2\sqrt {{2^{\cos e{c^{ - {1_t}}}}}} }}{2^{\cos e{c^{ - {1_t}}}}}\log 2.\frac{{ - 1}}{{x\sqrt {{x^2}} - 1}}\) \(\frac{{dy}}{{dt}} = \frac{1}{{2\sqrt {{2^{se{c^{ - {1_t}}}}}} }}{2^{se{c^{ - {1_t}}}}}\log 2.\frac{{ - 1}}{{x\sqrt {{x^2}} - 1}}\)…
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