Finding the partial derivative I think

Quote:

Originally Posted by ghow90

I think I'm supposed to find the partial derivative of the equation with respect to x and then with respect to y, right?

The Problem is...

Find the slope of the line that is parallel to the (a)xz-plane (b)yz-plane and tangent to the surface $z=xln(x+y^2)$ at the point $P(e,0,e)$

I just don't know how to find the partial derivatives of $z=xln(x+y^2)$

Differentiate z(x,y) wrt x while considering y as a constant:

Use the product rule: $\frac{\partial z}{\partial x} = \ln(x+y^2) \cdot 1 + x \cdot \frac1{x+y^2} \cdot 1 = \ln(x+y^2)+\frac x{x+y^2}$

Differentiate z(x,y) wrt y while considering x as a constant:

Keep in mind: x is a constant factor here. Use the chain rule: $\frac{\partial z}{\partial y} = x \cdot \frac1{x+y^2} \cdot 2y = \frac{2 x y}{x+y^2}$