Question

g(x; y) = 5(
28-2x^2- 4y)^2)^1/2
find partial derivative of g in terms of x any

Answer 1

\[g(x:y)= 5\sqrt{28-2x^{2} - 4y^{2}:}\]

Answer 2

\[g(x,y)=(28-2x ^{2}-4y ^{2})^{1/2}\]
\[g _{x}=\frac{ 1 }{ 2 }*5*(28-2x ^{2}-4y ^{2})^{-1/2}*-4x\]

Answer 3

that is what i thought initially but then a thought came into my mind when partially differentiating a number doesn't it become zero?

so when you differentiate (28-2x^2-4y^2) ^1/2 doesn't 1/2 become zero?

Answer 4

why become zero?

Answer 5

in writting an answer to u i have figured it out. differentiating 5x^2 wrt x does not give zero becoz 2 is a number it gives 10x

Answer 6

thx