Multivariable calculus

Partial Derivatives of e^x sin(y)

For f(x, y) = e to the x times sin(y), the partial derivative with respect to x is e to the x times sin(y), the function itself, and the partial derivative with respect to y is e to the x times cos(y). The exponential is its own derivative, so only the trigonometric factor changes.

f(x,y)=exsin⁡yf(x,y) = e^{x}\sin y
∂f∂x=exsin⁡y\frac{\partial f}{\partial x} = e^{x}\sin y
∂f∂y=excos⁡y\frac{\partial f}{\partial y} = e^{x}\cos y

Two one variable derivatives, nothing more

The function splits as a product where each factor owns one variable. Holding yy fixed makes sin⁡y\sin y a constant, and ddxex=ex\frac{d}{dx}e^{x} = e^{x}, so the xx partial reproduces the original function.

∂f∂x=sin⁡y⋅ex=f(x,y)\frac{\partial f}{\partial x} = \sin y \cdot e^{x} = f(x,y)

Holding xx fixed makes exe^{x} a constant multiplier and turns the sine into a cosine.

∂f∂y=excos⁡y\frac{\partial f}{\partial y} = e^{x}\cos y

At the origin, ∇f(0,0)=(0,1)\nabla f(0,0) = (0, 1). The tangent plane there is z=yz = y, so near the origin the surface looks like a plane tilted only in the yy direction.

The mistake: differentiating the exponent instead of the exponential

Because exe^{x} often appears as eg(x)e^{g(x)} in earlier courses, students reach for a chain rule factor that is not there. Here the exponent is simply xx, so the inside derivative is 11 and ∂∂xex=ex\frac{\partial}{\partial x}e^{x} = e^{x}.

  • A second slip is writing ∂f∂y=0\frac{\partial f}{\partial y} = 0 on the grounds that the exponential does not contain yy. The sin⁡y\sin y factor does, and it is what gets differentiated.
  • A third is losing the exe^{x} from the yy partial. It is a constant with respect to yy, and constant multipliers survive differentiation.

Why this function is special

Take second derivatives. From ∂f∂x=exsin⁡y\frac{\partial f}{\partial x} = e^{x}\sin y you get ∂2f∂x2=exsin⁡y\frac{\partial^2 f}{\partial x^2} = e^{x}\sin y, and from ∂f∂y=excos⁡y\frac{\partial f}{\partial y} = e^{x}\cos y you get ∂2f∂y2=−exsin⁡y\frac{\partial^2 f}{\partial y^2} = -e^{x}\sin y. They cancel.

∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0

A function with zero Laplacian is called harmonic, and harmonic functions model steady state temperature, electrostatic potential in charge free regions, and ideal fluid flow. This particular one is the imaginary part of the complex exponential ex+iye^{x + iy}, which is where its clean structure comes from.

Frequently asked questions

Can a partial derivative equal the original function?

Yes, and here ∂f∂x=f\frac{\partial f}{\partial x} = f. That is the defining property of the exponential in the xx direction, and it means the surface grows at a rate proportional to its own height as you move in xx.

What is the harmonic partner of this function?

The real part of ex+iye^{x+iy}, namely excos⁡ye^{x}\cos y. The pair satisfies the Cauchy Riemann equations, which say the xx partial of one equals the yy partial of the other up to sign.