Question

If e^xy=4 then what is dy over dx at the point (1, ln4)

Answer 1

`e^(xy)=4`

`(\mathrm d)/(\mathrm dx)e^(xy)=(\mathrm d)/(\mathrm dx)4`

`e^(xy)(\mathrm d)/(\mathrm dx)[xy]=0`

`e^(xy)\left(y+x(\mathrm dy)/(\mathrm dx)\right)=0`

`(\mathrm dy)/(\mathrm dx)=-\frac yx`

At the point
`(1,\ln4)`, the derivative takes on the value of
`-\frac{\ln4}1=-\ln4`.

Answer 2

-1.386

Explanation:

Given the function e^xy = 4 to differentiate the function with respect to x means differentiating the function implicitly since the given equation is a function of two variables x and y.

The equation e^xy = 4 can also be expressed as;

e^xy-4 = 0

Differentiating the function e^xy implicitly with respect to x using chain rule, let z = e^xy

Let u = xy...(1)

the equation becomes;

z = e^u...(2)

From equation 1; du/dx = xdy/dx + y (using product rule and differentiating implicitly)

from equation 2; dz/du = e^u

dz/dx = dz/du × du/dx

dz/dx = (xdy/dx + y)e^u

Since u = xy

dz/dx = e^xy(xdy/dx + y)

Substituting the resulting differential in the original equation, it will become;

e^xy(xdy/dx + y) = 0... (3) (note that differential of a constant gives zero)

Dividing both sides of equation 3 by e^xy gives;

xdy/dx+y = 0

xdy/dx = -y

dy/dx = -y/x

dy/dx at the point (1,ln4) gives;

dy/dx = -ln4/1

dy/dx = -ln4

dy/dx at (1,ln4) = -1.386