Differenciation help plz!!!!!!!!!! error free i think :)

implicit differentiation :
for the implicit function obtain dy/dx in terms of x and y.hence find the gradient at p (1, 2)

a) 1/x - 1/y = 4 (method 1 )
x^-1 - y^-1 = 4
-1x^-2 +1y^-2dy/dx =0
1/y^2dydx =1/x^2
dy/dx= y^2/x^2 (correct answer)
at p (1, 2) dy/dx= 4/1
= 4


b) 1/x -1/y = 4 (method 2)
y - x = 4xy (multiplying by xy)
dy/dx -1 = 4x dy/dx + 4y
(4x-1)dy/dx=-1-4y
dy/dx=(-1-4y)/(4x-1) (suppose to be correct)
at p(1 , 2) dy/dx=-9/3
=-3

my question is why is the derivative of method 1 not equal to method 2 plzzzzz provide me with an answer as to why it is wrong or the correct answer if i made an error

thank you in advance :)

Hi Jamal,

Your differentiation methods are spot on!

Consider rearranging the equation:

y - x = 4xy.

y - 4xy = x.

y (1 - 4x) = x.

y = x / (1 - 4x).

[1] y / x = 1 / (1 - 4x).

And again:

y - x = 4xy.

y = x + 4xy.

y = x (1 + 4y).

[2] y / x = (1 + 4y).

Combining [1] and [2]:

(y / x) * (y / x) = ((1 / (1 - 4x)) * (1 + 4y) = (1 + 4y) / (1 - 4x).

Or:

y² / x² = (1 + 4y) / (1 - 4x).

i.e. method one = method two.

However, you're not considering the point p.

Try substituting it in:

1 / x - 1 / y = 4.

1 / 1 - 1 / 2 = 4.

1 / 2 = 4.

This is nonsense. The fact is, that point p does not satisfy the equation of the line, hence the gradient at this point is not consistent.

You could infer this yourself:

If the point p satisfies the equation of the line, these two methods are equal when evaluated at p. The two methods give different answers. Hence, p does not satisfy the equation of the line.