Implicit Differentiation?

Question

x^2(cosy)+sin2y=xy?

Answer 1

Differentiating wrt x

2x . cosy + x^2 . (-Siny) . dy/dx + Cos2y . 2 . dy/dx = 1 . y + x . dy/dx

2xCosy - x^2 Siny . dy/dx + 2Cosy dy/dx = y + xdy/dx

Now solve for dy/dx

1.e^y + x . e^y . dy/dx = -1 . y^-2 . dy/dx

e^y + xe^y dy/dx = -1 / y^2 .dy/dx

Now solve for ....

Answer 2

x²·cos(y) + sin(2·y) = x·y

d/dx [ x²·cos(y) + sin(2·y) ] = d/dx [ x·y ]

As y is treated like a function of x during implicit differentiation, that makes x² times cos(y) a product of functions of x. Thus, use the product rule. The same applies for x times y.

x² · d/dx [ cos(y) ] + cos(y) · d/dx [ x² ] + d/dx [ sin(2·y) ] = x · d/dx [ y ] + y · d/dx [ x ]

Don't forget the chain rule with y, as it is a function itself, whose derivative is dy/dx :

x²·[-sin(y)]·dy/dx + cos(y)·(2·x) + cos(2·y)·(2)·dy/dx = x·(1)·dy/dx + y·(1)

It's a bit easier to write dy/dx as just y':

x²·[-sin(y)]·y' + cos(y)·(2·x) + cos(2·y)·(2)·y' = x·(1)·y' + y·(1)

-x²·sin(y)·y' + 2·x·cos(y) + 2·cos(2·y)·y' = x·y' + y

You want to solve for dy/dx (y'), so move all terms with that as part of them to one side and everything else to the other side (just like you would with a normal variable):

-x²·sin(y)·y' + 2·cos(2·y)·y' - x·y' = y - 2·x·cos(y)

Factor out the y':

y' · [ -x²·sin(y) + 2·cos(2·y) - x ] = y - 2·x·cos(y)

Divide to isolate it:

y' = [ y - 2·x·cos(y) ] / [ -x²·sin(y) + 2·cos(2·y) - x ]

Optionally, multiply by -1/-1 to simplify a bit:

y' = [ 2·x·cos(y) - y ] / [ x²·sin(y) - 2·cos(2·y) + x ]

——————————————————————————————————————

x·e^(y) = y - 1

d/dx [ x·e^(y) ] = d/dx [ y - 1 ]

Again, product rule on the left:

x · d/dx [ e^(y) ] + e^(y) · d/dx [ x ] = d/dx [ y ] - d/dx [ 1 ]

x·e^(y)·y' + e^(y)·(1) = y' - 0

x·e^(y)·y' + e^(y) = y'

Solve for y' now:

e^(y) = y' - x·e^(y)·y'

e^(y) = y' · [ 1 - x·e^(y) ]

e^(y)/[ 1 - x·e^(y) ] = y'