Differentiation Question 299
Question: If $ x=y\sqrt{1-y^{2},} $ then $ \frac{dy}{dx}= $
[MP PET 2001]
Options:
A) 0
B) x
C) $ \frac{\sqrt{1-y^{2}}}{1-2y^{2}} $
D) $ \frac{\sqrt{1-y^{2}}}{1+2y^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ x=y\sqrt{1-y^{2}} $
Differentiate with respect to x, $ 1=\frac{dy}{dx}\sqrt{1-y^{2}}+y.\frac{1}{2\sqrt{1-y^{2}}}.(-2y).\frac{dy}{dx} $
Therefore $ 1=\frac{dy}{dx}\sqrt{1-y^{2}}-\frac{y^{2}}{\sqrt{1-y^{2}}}.\frac{dy}{dx} $
Therefore $ 1=\frac{dy}{dx}[ \frac{1-y^{2}-y^{2}}{\sqrt{1-y^{2}}} ] $
Therefore $ 1=\frac{dy}{dx}[ \frac{1-2y^{2}}{\sqrt{1-y^{2}}} ] $
$ \frac{dy}{dx}=\frac{\sqrt{1-y^{2}}}{1-2y^{2}} $ .
BETA
