Advanced Partial Derivatives: Higher-Order, Differentiability & Laplace Equation

To find the partial derivative with respect to y, we treat x as a constant. The first term is x³ * tan(y). Since x³ is a constant coefficient, we only differentiate tan(y), which becomes sec²(y). The second term, 4x, contains no y variable, so it is treated as a constant and its derivative is 0. Therefore, the result is x³ * sec²(y).

The partial derivative with respect to x measures the rate of change of the function as x changes while y is held fixed. Geometrically, holding y fixed at the value b corresponds to slicing the surface z = f(x, y) with the vertical plane y = b. This intersection creates a curve (a trace) on that plane. The partial derivative f_x(a, b) represents the slope of the tangent line to this specific trace curve at the point where x = a.

To find the partial derivative with respect to z, we treat both x and y as constants. The function is a product of the constant term (x² * y) and the variable term (z³). We apply the power rule to z³, which gives us 3z². We then multiply this result by the constant coefficient (x² * y). This yields 3 * x² * y * z².

The partial derivative f_x(1,2) is the instantaneous rate at which f changes with x while y is held fixed; by definition it is the limit of the ratio (change in f)/(change in x) as the change in x goes to zero. To estimate that from the table, take two function values at the same y = 2 that lie the same distance to the left and right of x = 1. Those values are f(0.8,2) = 3.2 and f(1.2,2) = 4.4. The average rate of change of f with respect to x over the interval from x = 0.8 to x = 1.2 is (4.4 − 3.2)/(1.2 − 0.8) = 1.2/0.4 = 3.0. Because these two sample points straddle x = 1 and are equally spaced about it, this ratio gives a good approximation to the instantaneous rate f_x(1,2), assuming f is reasonably smooth near (1,2). Thus the best estimate from the table is 3.0.

In multivariable calculus, differentiability is a stronger condition than the mere existence of partial derivatives. Just like in single-variable calculus, if a function is differentiable at a point (meaning it has a valid local linear approximation or tangent plane), it implies that the graph has no holes or jumps at that point. Therefore, differentiability guarantees continuity. However, the reverse is not true; a function can be continuous but have a sharp point (like a cone), making it non-differentiable.

For f(x,y) = sin(xy), we need the chain rule for partial derivatives. For ∂f/∂x, we treat y as constant. The derivative of sin(u) with respect to x is cos(u)∂u/∂x where u = xy. Since ∂u/∂x = y (treating y constant), we get ∂f/∂x = cos(xy)y = ycos(xy). Similarly for ∂f/∂y, we treat x as constant. The derivative of sin(u) with respect to y is cos(u)∂u/∂y where u = xy. Since ∂u/∂y = x (treating x constant), we get ∂f/∂y = cos(xy)x = xcos(xy). Option A forgets the chain rule factor. Option C swaps the factors. Option D confuses sine with cosine in the differentiation.

The partial derivative f_x represents the rate of change of altitude as you move in the positive x-direction (East). Since walking East causes you to go downhill, the altitude is decreasing, so f_x must be negative. The partial derivative f_y represents the rate of change of altitude as you move in the positive y-direction (North). Since walking North keeps you at the same altitude, the rate of change is zero, so f_y is zero.

A partial derivative examines how a function changes when exactly one variable changes and all others are held fixed. The notation ∂P/∂V indicates that we are differentiating the Pressure function P with respect to the variable V (Volume). This implies that the other variable, T (Temperature), is treated as a constant. Therefore, this derivative represents the rate at which Pressure changes as Volume changes, while Temperature remains constant (an isothermal process).

We need to find ∂f/∂x and then evaluate it at (0,π). Step 1: Compute ∂f/∂x. For f(x,y) = cos(x)*sin(y), we treat y as constant (so sin(y) is constant) and differentiate with respect to x. The derivative of cos(x) with respect to x is -sin(x). Therefore, ∂f/∂x = -sin(x)*sin(y). Step 2: Evaluate at (0,π). Substitute x = 0 and y = π into the partial derivative: ∂f/∂x(0,π) = -sin(0)sin(π). Since sin(0) = 0 and sin(π) = 0, we get -00 = 0. The result is 0. Option B would be incorrect because sin(0) = 0 regardless of sin(π). Option C incorrectly evaluates sin(0) as 1. Option D is wrong because the function and its partial derivative are well-defined everywhere.

To differentiate with respect to y, we treat x as a constant. We can rewrite the function as x * (x + y)^(-1). Using the chain rule (or the general power rule), we bring down the exponent -1 to get -1 * x * (x + y)^(-2). Then we multiply by the derivative of the inside function (x + y) with respect to y, which is just 1. This gives us -x * (x + y)^(-2), which can be rewritten as the fraction -x / (x + y)². Alternatively, using the quotient rule: (Low * dHigh - High * dLow) / Low². Here dHigh (derivative of x w.r.t y) is 0, and dLow (derivative of x+y w.r.t y) is 1. So, ((x+y)*0 - x*1) / (x+y)² = -x / (x+y)².

To find the partial derivative with respect to z, we treat x and y as constants. This means the term x * e^(y) acts as a constant coefficient. We only need to differentiate sin(z) with respect to z. The derivative of sin(z) is cos(z). Multiplying this by the constant coefficient x * e^(y) gives us the result x * e^(y) * cos(z).

We compute the partial derivatives. ∂V/∂r = 2πrh and ∂V/∂h = πr². The rate of change with respect to radius depends on the current height (and radius), and the rate of change with respect to height depends on the square of the radius. Therefore, you cannot say one is always more sensitive than the other without knowing the specific values of r and h; the sensitivity is dynamic based on the shape of the cylinder.

The second-order partial derivative ∂²f/∂x² means we take the partial derivative with respect to x twice. First, find the first partial derivative with respect to x, treating y as a constant: ∂f/∂x = 3x²e^(y). Now, take the partial derivative of this result with respect to x again, still treating y as a constant: the derivative of 3x² is 6x, and e^(y) remains as a constant multiplier. Therefore, ∂²f/∂x² = 6x * e^(y) = 6xe^(y).

For each function, compute the sum of the second partial derivatives. For u = x²+y², ∂²u/∂x² = 2, ∂²u/∂y² = 2, sum = 4 ≠ 0. For u = e^x sin(y), ∂²u/∂x² = e^x sin(y), ∂²u/∂y² = -e^x sin(y), sum = 0. For u = x²-y², ∂²u/∂x² = 2, ∂²u/∂y² = -2, sum = 0. For u = ln(x²+y²), after computing (as above), the sum is 0. Therefore, only option A does not satisfy Laplace's equation.