Differentiation Question 82
If $ f(u)=f(x,y,z) $ be a homogeneous function of degree $ n $ in $ x,y,z $ then $ x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}+z\frac{\partial f}{\partial z}= $
Options:
A) $\nu$
B) $ nF(u) $
C) $ \frac{nF(u)}{{F}’(u)} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Since $ F(u) $ is homogeneous of degree n in $ x,y,z $. $ \therefore $ $ x\frac{\partial }{\partial x}(F(u))+y.\frac{\partial }{\partial y}(F(u))+z\frac{\partial }{\partial z}(F(u))=nF(u) $
Therefore $ x.{F}’(u)\frac{\partial u}{\partial x}+y{F}’(u)\frac{\partial u}{\partial y}+z{F}’(u)\frac{\partial u}{\partial z}=nF(u) $
Therefore $ \frac{{{\partial }^{2}}z}{\partial x^{2}}=a^{2}{{\sec }^{3}}(y-ax)+a^{2}\sec (y-ax){{\tan }^{2}}(y-ax) $ .
BETA
