Differentiation Question 72
Question: If $ z={{\tan }^{-1}}( \frac{x}{y} ) $ , then $ z _{x}:z _{y}= $
Options:
A) $ y:x $
B) $ x:y $
C) $ -y:x $
D) $ -x:y $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{\partial z}{\partial x}=\frac{1}{1+\frac{x^{2}}{y^{2}}}.\frac{1}{y}=\frac{y}{x^{2}+y^{2}} $
$ \frac{\partial z}{\partial y}=\frac{1}{1+\frac{x^{2}}{y^{2}}}.( -\frac{x}{y^{2}} )=-\frac{x}{x^{2}+y^{2}} $
$ \therefore $ $ \frac{\partial z}{\partial x}:\frac{\partial z}{\partial y}=y:-x $ i.e., $ z _{x}:z _{y}=-y:x $
BETA
