Implicit differentiation

[Cadbury]

Question is: Find dy/dx in both cases:

a) e^xy = 5

b) sin2xcosy = 3

Please help!

[krtxmrtz]

In the first case, if you define:

f(x,y) = e^xy then, f = 5 and df = 0 (df is the differential). Using the symbol d_p meaning 'partial derivative' then:

df = (d_p/dx) dx + (d_p/dy) dy

from which,

dy/dx = -(d_p/dx) / (d_p/dy) = -ye^xy/xe^xy = -y/x

In the second case you do just the same thing:

f = sin2xcosy

dy/dx = -(d_p/dx) / (d_p/dy) = -(2cos2xcosy) / (-sin2xsiny) = 2 / (tan2xtany)

By the way, the latter function can't be right or, rather, has no complying point (x,y) for both the sine and cosine vary from -1 to 1 so a product of a sine and a cosine can't be equal to 3.

[Cadbury]

That seems to make confusing sense. So are you saying dy/dx of e^xy =- y/x?

The sin and cos one makes sense.

Thank you for your help by the way.

[krtxmrtz]

That seems to make confusing sense. So are you saying dy/dx of e^xy =- y/x?
...

That's right.

From e^xy = 5 you get:

xy = ln(5)
y = ln(5) / x

and dy/dx = -ln(5) / x²

i.e. dy/dx = -y/x

[Glaysher]

Question is: Find dy/dx in both cases:

a) e^xy = 5

b) sin2xcosy = 3

Please help!

a)

Make t = xy then dt/dx = y + x dy/dx

And e^t differentiates to e^t

Then d/dx(e^xy) = e^t(y + x dy/dx) = e^xy(y + x dy/dx)

5 differentiates to 0 so we get:

e^xy(y + x dy/dx) = 0

y + x dy/dx = 0

x dy/dx = -y

dy/dx = -y/x

[Glaysher]

Question is: Find dy/dx in both cases:

a) e^xy = 5

b) sin2xcosy = 3

Please help!

b)

By product rule

d/dx(sin 2x cos y) = 2 cos 2x cos y - sin 2x sin y dy/dx

3 differentiates to zero so

2 cos 2x cos y - sin 2x sin y dy/dx = 0

sin 2x sin y dy/dx = 2 cos 2x cos y

dy/dx = (2 cos 2x cos y)/(sin 2x sin y) = 2 cot 2x cot y