Partials of ln(3 + sin(x)cos(y))

Logarithm outside, product inside

Write $u = 3 + \sin x\cos y$. The logarithm rule gives $\frac{\partial f}{\partial x} = \frac{1}{u}\frac{\partial u}{\partial x}$, so the whole job reduces to differentiating the inside, which you already know how to do.

Divide each by $u$ and you have both answers. The constant $3$ contributes nothing to the numerators, since the derivative of a constant is zero, but it stays in the denominator because that denominator is the original inside function.

At the origin $f = \ln 3 \approx 1.0986$ and the gradient is $\left(\frac{1}{3}, 0\right)$.

The mistake: dropping the 3 from the denominator

The single most common error is writing the denominator as $\sin x\cos y$, on the reasoning that the $3$ differentiates away. It does differentiate away in the numerator, but the denominator is $u$ itself, untouched by differentiation.

• The chain rule for $\ln u$ is $\frac{u'}{u}$, and $u$ in the denominator is the full inside expression.
• A second slip is splitting the logarithm as $\ln 3 + \ln(\sin x\cos y)$. There is no such rule for a sum inside a logarithm, and $\sin x \cos y$ goes negative anyway.
• A third is losing the minus sign in the $y$ partial. It comes from $\frac{d}{dy}\cos y = -\sin y$.

Why the function is defined everywhere

Both $\sin x$ and $\cos y$ live between $-1$ and $1$, so their product lives between $-1$ and $1$ and the inside $3 + \sin x\cos y$ lives between $2$ and $4$. A logarithm needs a positive argument, and $2$ is comfortably positive, so there is no domain restriction to worry about and no denominator that can vanish.

Critical points need both numerators to be zero at once. From $\cos x\cos y = 0$ and $\sin x\sin y = 0$ you cannot have $\cos x = 0$ and $\sin x = 0$ together, so the solutions pair $\cos x = 0$ with $\sin y = 0$, or $\cos y = 0$ with $\sin x = 0$. Those are the points where $\sin x \cos y$ hits $\pm 1$ or $0$, that is the peaks, valleys, and saddles of the surface.