Differentiation Question 78
Question: If $ u={{\sin }^{-1}}( \frac{y}{x} ), $ then $ \frac{\partial u}{\partial x} $ is equal to
[Tamilnadu (Engg.) 1992]
Options:
A) $ -\frac{y}{x^{2}+y^{2}} $
B) $ \frac{x}{\sqrt{1-y^{2}}} $
C) $ \frac{-y}{\sqrt{x^{2}-y^{2}}} $
D) $ \frac{-y}{x\sqrt{x^{2}-y^{2}}} $
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Answer:
Correct Answer: D
Solution:
$ u={{\sin }^{-1}}\frac{y}{x} $ ; $ \tan u $
$ \frac{\partial u}{\partial x}=\frac{1}{\sqrt{1-\frac{y^{2}}{x^{2}}}}.( -\frac{y}{x^{2}} )=-\frac{y}{x\sqrt{x^{2}-y^{2}}} $ .
BETA
