Evaluating Partial Derivatives at a Point

From a Partial Derivative to Its Value at a Point

A partial derivative is itself a function. Finding a formula such as \(f_x(x,y)\) tells us how the function changes with respect to \(x\), with \(y\) held fixed, at any point where that partial derivative exists. Evaluating it at a point means taking that formula and substituting the point’s coordinates.

Keeping these two steps separate matters. The function value \(f(a,b)\), the partial derivative \(f_x(a,b)\), and the partial derivative \(f_y(a,b)\) are generally different numbers. Each answers a different question.

Definition: If \(f\) is a function of \(x\) and \(y\), then \(f_x(a,b)\) is the value of its partial derivative with respect to \(x\) at \((a,b)\), and \(f_y(a,b)\) is the value of its partial derivative with respect to \(y\) there. To calculate either value, first obtain the corresponding partial derivative function, then evaluate it at \(x=a,\ y=b\). The point must be in the domain of \(f\), and the relevant partial derivative must exist there.

A Reliable Evaluation Procedure

For a function \(f(x,y)\) and a requested point \((a,b)\), write down the two coordinates clearly. Differentiate with respect to the requested variable while treating the other variable as fixed. Only after obtaining the partial derivative formula should you substitute \(x=a\) and \(y=b\). Keep the resulting values labelled: for example, \(f_x(a,b)\) and \(f_y(a,b)\).

Check the point.
Confirm that the original formula is defined at the specified coordinates. In particular, a rational expression requires a nonzero denominator.

Differentiate symbolically.
Find the requested partial derivative as a function of the variables, holding all other variables fixed.

Substitute the coordinates.
Replace each variable in the derivative formula with its coordinate and simplify the result.

Label and check.
State which partial derivative was evaluated and verify the substitution and arithmetic.

The derivative formula is evaluated at the point in the same way as any other function formula: its input is the coordinate pair. There is no additional differentiation at this stage.

Why Differentiation Comes Before Substitution

The partial derivative with respect to \(x\) at \((a,b)\) is defined by varying \(x\) while keeping \(y=b\) fixed. When the limit exists, its definition is

f_x(a,b)=\lim_{h\to 0}\frac{f(a+h,b)-f(a,b)}{h}.

For a general point \((x,y)\), the same definition produces the derivative function \(f_x(x,y)\), when the corresponding limits exist. Evaluating that function at \((a,b)\) sets \(x=a\) and \(y=b\), giving the limit above. This is why substituting into the derivative formula gives the partial derivative at the specified point. The limit must exist, and the function must be defined where the difference quotient is evaluated.

Substituting both coordinates into \(f\) before differentiating changes the task: it reduces the function to a single number, leaving no variable to differentiate. The next example illustrates the distinction.

Worked Example: Evaluating Polynomial Partial Derivatives

Let \(f(x,y)=x^2y+3xy^2-y^3\). Find \(f_x(1,-1)\), \(f_y(1,-1)\), and compare them with \(f(1,-1)\).

Differentiate with respect to \(x\), holding \(y\) fixed. The term \(x^2y\) gives \(2xy\), the term \(3xy^2\) gives \(3y^2\), and the term \(-y^3\) is constant with respect to \(x\):

$$ f_x(x,y)=2xy+3y^2. $$

Now substitute \(x=1\) and \(y=-1\):

$$ f_x(1,-1)=2(1)(-1)+3(-1)^2=-2+3=1. $$

Next differentiate with respect to \(y\), holding \(x\) fixed:

$$ f_y(x,y)=x^2+6xy-3y^2. $$

Evaluate at the same point:

$$ f_y(1,-1)=(1)^2+6(1)(-1)-3(-1)^2 =1-6-3=-8. $$

Finally, evaluate the original function to distinguish its value from the partial derivative values:

$$ f(1,-1)=(1)^2(-1)+3(1)(-1)^2-(-1)^3 =-1+3+1=3. $$

Thus \(f_x(1,-1)=1\), \(f_y(1,-1)=-8\), and \(f(1,-1)=3\). The values are different because they represent different quantities. As a check, differentiating each displayed partial derivative formula gives \(f_{xx}=2y\) and \(f_{yy}=6x-6y\); these equal \(-2\) and \(12\) at the point, respectively, consistent with the polynomial formulas. These extra derivatives are not needed to obtain the requested first partials.

Evaluating Partial Derivatives of a Rational Function

For a quotient, differentiate the formula while the variables are still present, then substitute. The original denominator must be nonzero at the point for the function to be defined there. In a quotient-rule calculation, it is also important to differentiate both numerator and denominator with respect to the chosen variable.

Worked Example: A Rational Function at a Specified Point

Let \(f(x,y)=\dfrac{x^2+3y}{x-y}\). Find \(f_x(2,0)\) and \(f_y(2,0)\).

At the point, the denominator is \(2-0=2\ne 0\), so the function is defined. Write the numerator as \(N=x^2+3y\) and the denominator as \(D=x-y\). For the \(x\)-partial, \(N_x=2x\) and \(D_x=1\). The quotient rule gives

f_x(x,y)=\frac{(2x)(x-y)-(x^2+3y)(1)}{(x-y)^2} =\frac{x^2-2xy-3y}{(x-y)^2}.

Substitute the coordinates into this derivative formula:

f_x(2,0)=\frac{(2)^2-2(2)(0)-3(0)}{(2-0)^2} =\frac{4}{4}=1.

For the \(y\)-partial, \(N_y=3\) and \(D_y=-1\). Therefore

f_y(x,y)=\frac{3(x-y)-(x^2+3y)(-1)}{(x-y)^2} =\frac{3x-3y+x^2+3y}{(x-y)^2} =\frac{x^2+3x}{(x-y)^2}.

Evaluating at \((2,0)\) gives

f_y(2,0)=\frac{(2)^2+3(2)}{(2-0)^2} =\frac{4+6}{4} =\frac{5}{2}.

As a check on the quotient-rule expressions, differentiating \(f=N/D\) with respect to \(x\) requires \(N_xD-ND_x\), while differentiation with respect to \(y\) requires \(N_yD-ND_y\). Substituting the point into these unsimplified numerators gives \(4\) for the \(x\)-partial numerator and \(10\) for the \(y\)-partial numerator; in both cases the squared denominator is \(4\). The resulting values, \(1\) and \(5/2\), agree with the simplified formulas.

Evaluating Trigonometric Partial Derivatives

Trigonometric expressions follow the same differentiate-then-substitute procedure. The usual calculus convention is that angles in trigonometric functions are measured in radians. If the derivative contains a trigonometric value at a familiar angle, simplify it exactly before using a decimal approximation.

Worked Example: Trigonometric Partials at Exact Angles

Let \(f(x,y)=\sin(xy)+y\cos x\). Find both first partial derivatives at \(\left(\frac{\pi}{2},1\right)\).

For \(f_x\), hold \(y\) fixed. Applying the chain rule to \(\sin(xy)\) gives \(y\cos(xy)\), and differentiating \(y\cos x\) gives \(-y\sin x\):

f_x(x,y)=y\cos(xy)-y\sin x.

Now substitute \(x=\frac{\pi}{2}\) and \(y=1\):

f_x\left(\frac{\pi}{2},1\right) =1\cos\left(\frac{\pi}{2}\right)-1\sin\left(\frac{\pi}{2}\right) =0-1=-1.

For \(f_y\), hold \(x\) fixed. The derivative of \(\sin(xy)\) with respect to \(y\) is \(x\cos(xy)\), and that of \(y\cos x\) is \(\cos x\):

f_y(x,y)=x\cos(xy)+\cos x.

At the point,

f_y\left(\frac{\pi}{2},1\right) =\frac{\pi}{2}\cos\left(\frac{\pi}{2}\right) +\cos\left(\frac{\pi}{2}\right) =0+0=0.

As a check, the chain-rule factors are \(y\) for differentiating \(xy\) with respect to \(x\), and \(x\) for differentiating it with respect to \(y\). At the specified point those factors are \(1\) and \(\frac{\pi}{2}\), exactly the factors appearing in the evaluated expressions. The exact results are \(f_x\left(\frac{\pi}{2},1\right)=-1\) and \(f_y\left(\frac{\pi}{2},1\right)=0\).

A Common Mistake: Substituting Too Early

A tempting but incorrect approach is to replace the variables with their coordinate values in the original function before differentiating. For example, take \(g(x,y)=x^2y\) and the point \((2,3)\). Substituting both coordinates first gives \(g(2,3)=12\), a constant. Differentiating that number would give \(0\), but this does not calculate either partial derivative.

The correct calculation keeps the variables until after differentiation:

g_x(x,y)=2xy, \qquad g_y(x,y)=x^2.

Then evaluate each formula at the point:

g_x(2,3)=2(2)(3)=12, \qquad g_y(2,3)=(2)^2=4.

The incorrect early substitution discarded the information about how the function changes when one coordinate varies and the other stays fixed. The values \(12\) and \(4\) are the two different coordinatewise rates at \((2,3)\), not the function value. Indeed, \(g(2,3)=12\) happens to match \(g_x(2,3)\) here, but that numerical coincidence does not make them the same quantity.

Key takeaway: A partial derivative formula is a function of the coordinates. Differentiate while the variables are present; then substitute the point into the derivative formula. Do not confuse the resulting partial derivative value with the original function value.

Check Your Understanding

For each question, differentiate first and then evaluate at the given point. Keep the function value separate from the partial derivative values.

For \(f(x,y)=x^2+xy\), find \(f_x(1,2)\) and \(f_y(1,2)\).
For \(g(x,y)=\dfrac{x+y}{x-y}\), is \(g\) defined at \((1,1)\)? Explain before attempting to evaluate its partial derivatives there.
For \(h(x,y)=\sin x+y^2\), find \(h_x(0,2)\) and \(h_y(0,2)\).
For \(p(x,y)=xy^2\), explain why substituting \((3,2)\) into \(p\) before differentiating cannot produce \(p_x(3,2)\).
For \(q(x,y)=x^2y\), calculate \(q(2,1)\), \(q_x(2,1)\), and \(q_y(2,1)\). Which values represent partial derivatives?