# Equation of tangent - Implicit or Partial DifferentiatioN?

1. Jun 11, 2012

### dan38

1. The problem statement, all variables and given/known data
Need to find the tangent to the curve at: e^(xy) + x^2*y - (y-x)^2 + 3

I just implicitly differentiate the expression to find the gradient and then use the points given to find the equation, right?
Or does this involve partial differentiation?

2. Relevant equations

3. The attempt at a solution

2. Jun 11, 2012

### Infinitum

None of those

Just differentiate with respect to x, using the chain rule and product rule.

3. Jun 11, 2012

### ehild

Is it a curve e^(xy) + x^2*y - (y-x)^2 + 3=constant, or a surface F(x,y)=e^(xy) + x^2*y - (y-x)^2 + 3? And at which point do you need to find the tangent or gradient?

ehild

4. Jun 11, 2012

### dan38

curve at (0,2)
how do you differentiate with respects to "x" if there's a "y" without one of those methods..

5. Jun 11, 2012

### Infinitum

As I said, use product and chain rules. For example, the derivative of the first term would be,

$$\frac{de^{xy}}{dx} = e^{xy} (y + x\cdot \frac{dy}{dx})$$

6. Jun 11, 2012

### dan38

ah I see
would it have been wrong to implicit differentiation then?

7. Jun 11, 2012

### Infinitum

Implicit differentiation is used when you cannot differentiate the terms using these methods, specifically when you have to integrate function raised to another function of the same variable. This is not the case with the given curve, so there is no need for implicit differentiation.

8. Jun 12, 2012

### ehild

Infinitum, what you did, it is implicit differentiation.

y is given as a function of x implicitly by an equation R(x, y) = 0. We differentiate R(x, y) with respect to x and then with respect to y, multiplying it with dy/dx.

ehild

9. Jun 12, 2012

### HallsofIvy

Staff Emeritus
What, exactly, do you mean by "implicit differentiation"? It is clearly called for in this problem.

10. Jun 13, 2012

### Infinitum

You're right. I don't know, but absent-mindedly, I interchange definitions sometimes....:surprised

Apologies, dan38!