Question

If 16=2xy^2+x^2y find dy/dx(1,0)

Answer 1

Explanation:

`{ \rm{16 = {2xy}^(2) + {x}^(2y) }} \\`

- Introduce dy/dx on both sides;

`{ \tt{ (dy)/(dx) (16) = (dy)/(dx) (2 {xy}^(2)) + (dy)/(dx) ( {x}^(2y) )}} \\ \\ { \tt{0 = \: \: \: \: \: \: \: \: \: \: \: \: 2( {y}^(2) +(dy)/(dx) ) \: \: \: \: \: + \: \: \: (dy)/(dx) ( {x}^(2y) )}} \\ \\ { \tt{ \uparrow}} \\ { \tt{product \: rule}}`

- Let's differentiate dy/dx(x^2y) alone;

`y = {x}^(2y) \\ ln(y) = ln( {x}^(2y) ) \\ ln(y ) = 2y ln(x) \\ \\ (1)/(ydx) dy = 2 ln(x) (dy)/(dx) + (2y)/(x) \\ \\ (dy)/(dx) ( (1)/(y) - 2 ln(x) ) = (2y)/(x) \\ \\ (dy)/(dx) = \frac{ {2y}^(2) }{x(1 - 2 ln(x)) }`

- Therefore;

`{ \tt{0 = 2( {y}^(2) + (dy)/(dx)) + \frac{ {2y}^(2) }{x(1 - 2 ln(x) )} }} \\ \\ { \tt{ 2(dy)/(dx) = - 2 {y}^(2) - \frac{ {2y}^(2) }{x(1 - 2 ln(x)) } }} \\ \\ { \tt{ (dy)/(dx) = - {y}^(2)(1 + (1)/(x(1 - 2 ln(x) )) ) }}`

- For points (1, 0);

`{ \tt{ (dy)/(dx) = 0}} \\`

Answer 2

We have the following function

`16=2xy^2+x^2y`

if we derive the function with respect to x we get

`(d)/(dx)\mleft(y\mright)=(-2xy-2y^2)/(x\left(x+4y\right))`

if we evaluate the result function in point 1,0 we will obtain

`(-2\cdot1\cdot0-2\cdot0)/(1(1+4\cdot0))=0`

Thus, dy/dx at point 1,0 is 0