I am unsure of how to take the derivative of this equation. It may be the exponents that are giving me trouble but I'm not sure exactly.

Find the equation of the tangent line to the curve 4e^xy = 2x + y at point (0,4).

On the left side, is the "xy" the exponent of e, or do you mean 4(e^x)y ?

Use implicit differentiation to get the slope m = dy/dx at x=0, y=4.
If 4 e^(xy) = 2x + y, then
4 e^(xy)* (y + x dy/dx) = 2 + dy/dx
Solve for dy/dx = m. The tangent line equation is then
(y-4) = m x

the exponent of e is "xy".

It's must clearer now, thank you.

In that case, my derivation should be correct. Substitute x=0, y=4 in the equation
4 e^(xy)* (y + x dy/dx) = 2 + dy/dx
4*1*(4 + 0) = 2 + dy/dx
dy/dx = 14
y-4 = 14 x
y = 14 x + 4

To find the equation of the tangent line to the curve, we need to first find the derivative of the equation with respect to x using implicit differentiation.

The equation of the curve is given as 4e^xy = 2x + y.

To begin, we treat y as a function of x and differentiate both sides of the equation with respect to x.

On the left side, we have the product of two functions, 4 and e^xy. To differentiate this, we use the product rule:

d/dx (u * v) = u * v' + v * u'

Here, u = 4 and v = e^xy. The derivative of u with respect to x is 0 since it is a constant, and the derivative of v with respect to x can be found using the chain rule:

d/dx (e^u) = e^u * u'

Applying the product rule, we have:

d/dx (4e^xy) = 4 * d/dx (e^xy) + e^xy * d/dx (4)

Since d/dx (4) is 0, the equation becomes:

d/dx (4e^xy) = 4 * d/dx (e^xy)

Next, we differentiate the right side of the equation. The derivative of 2x with respect to x is 2, and the derivative of y with respect to x is dy/dx.

Now, substituting everything back into the original equation, we have:

4 * d/dx (e^xy) = 2 + dy/dx

To solve for dy/dx, we isolate it by moving the terms with dy/dx to one side of the equation:

4 * d/dx (e^xy) - dy/dx = 2

Factoring out dy/dx on the left side:

dy/dx * (4 - e^xy) = 2

Finally, we solve for dy/dx by dividing both sides by (4 - e^xy):

dy/dx = 2 / (4 - e^xy)

Now that we have the slope (dy/dx) at the point (0,4), we can substitute these values into the equation y-4 = m(x - 0), where m is the slope we just calculated.

Plugging in the values, we get:

y - 4 = 2 / (4 - e^(0*4))(x - 0)

Simplifying further:

y - 4 = 2 / (4 - 1)(x - 0)

y - 4 = 2 / 3 * x

Rearranging the equation, we get the equation of the tangent line:

y = (2/3)x + 4

So, the equation of the tangent line to the curve 4e^xy = 2x + y at the point (0,4) is y = (2/3)x + 4.