Multivariable calculus

Partial Derivatives of e^(sin x + cos y)

Q: Where are the critical points?

The exponential is never zero, so both partials vanish only where \cos x = 0 and \sin y = 0. That is x = \frac{\pi}{2} + k\pi paired with y = m\pi for integers k and m, an infinite lattice of isolated critical points.

Q: Does it matter that the exponent is a sum rather than a product?

It makes the work easier. A sum splits under partial differentiation, so \frac{\partial u}{\partial x} ignores \cos y completely. With a product like \sin x\cos y in the exponent you would carry the other factor through as a constant multiplier.

The partials of e^(sin x + cos y) are f_x = cos(x) times e^(sin x + cos y) and f_y = -sin(y) times e^(sin x + cos y). The exponential reproduces itself, and the inner partial of sin x + cos y is cos x for the x partial and -sin y for the y partial.

f(x,y) = e^{\sin x + \cos y}

\frac{\partial f}{\partial x} = \cos(x)\,e^{\sin x + \cos y}

\frac{\partial f}{\partial y} = -\sin(y)\,e^{\sin x + \cos y}

Two layers, taken one at a time

The outer layer is $e^u$ with $u = \sin x + \cos y$ . Since $\frac{d}{du}e^u = e^u$ , the exponential comes along unchanged. The inner layer separates cleanly: $x$ appears only in $\sin x$ and $y$ only in $\cos y$ , so each partial sees a single term.

\frac{\partial f}{\partial x} = e^{\sin x + \cos y}\cdot\cos x

\frac{\partial f}{\partial y} = e^{\sin x + \cos y}\cdot(-\sin y)

Because the variables separate inside the exponent, $f$ also factors as $e^{\sin x}\,e^{\cos y}$ . Differentiating that form gives the same two answers, which makes it a good independent check on your work.

The sign on the y partial, and a misplaced derivative

The $y$ partial is negative wherever $\sin y > 0$ , which covers the whole interval $0 < y < \pi$ . The common error is dropping the minus from the derivative of cosine. At $x = 0$ , $y = 1$ the true value is $-\sin(1)\,e^{\cos 1} \approx -1.444$ , and the sign error reports $+1.444$ .

A structurally different mistake is writing $e^{\cos x - \sin y}$ , differentiating the exponent in place and leaving the result up there. The chain rule multiplies by the inner derivative; it does not replace the inner function with it.

Bounds you can check an answer against

Both $\sin x$ and $\cos y$ live in $[-1, 1]$ , so the exponent lies in $[-2, 2]$ and $f$ lies in $[e^{-2},\, e^{2}]$ , roughly $0.135$ to $7.389$ . Since $|\cos x| \le 1$ , that forces $|f_x| \le e^2 \approx 7.389$ as well.

Any candidate partial that grows without bound is wrong before you check a single number. This kind of bound costs one line and catches whole classes of algebra slips.

Frequently asked questions

Where are the critical points?

The exponential is never zero, so both partials vanish only where $\cos x = 0$ and $\sin y = 0$ . That is $x = \frac{\pi}{2} + k\pi$ paired with $y = m\pi$ for integers $k$ and $m$ , an infinite lattice of isolated critical points.

Does it matter that the exponent is a sum rather than a product?

It makes the work easier. A sum splits under partial differentiation, so $\frac{\partial u}{\partial x}$ ignores $\cos y$ completely. With a product like $\sin x\cos y$ in the exponent you would carry the other factor through as a constant multiplier.