8.05d - First and second partial derivatives | OCR A-Level Further Mathematics A

Differentiate a function of two variables while holding one variable fixed. You will use both standard notations, calculate pure and mixed second derivatives, and show that a surface satisfies an identity involving its partial derivatives.

First partial derivatives

Can the same point on a surface have two different gradients? On $z=x^2+y^2$ , at $(1,2,5)$ , the answer is yes: the section parallel to the $x$ -direction has gradient $2$ , while the section parallel to the $y$ -direction has gradient $4$ . A gradient needs a direction as well as a point.

You already know

Fixing $y=b$ gives the section $z=f(x,b)$ ; fixing $x=a$ gives $z=f(a,y)$ . Use the ordinary power, product and chain rules on a one-variable function.

To find the first gradient, treat $y$ as a constant and differentiate with respect to $x$ . To find the second, treat $x$ as a constant and differentiate with respect to $y$ . Thus $xy^2$ has derivative $y^2$ with respect to $x$ , but $2xy$ with respect to $y$ .

First partial derivatives

For $z=f(x,y)$ , the first partial derivatives are

f_x=\frac{\partial f}{\partial x}=\frac{\partial z}{\partial x}, \qquad f_y=\frac{\partial f}{\partial y}=\frac{\partial z}{\partial y}.

In $f_x$ , hold $y$ fixed; in $f_y$ , hold $x$ fixed. For example, if $f=x^2+y^2$ , then $f_x=2x$ and $f_y=2y$ .

The symbol $\partial$ distinguishes these derivatives from an ordinary derivative of a function of one independent variable. In particular, $\partial y/\partial x=0$ here because $x$ and $y$ are independent coordinates. It does not say that an arbitrary curve $y=y(x)$ has zero derivative. This notation and the rule of holding the other variable constant are not in the formula booklet.

[DIAGRAM: asset_name: Two slopes at one surface point; asset_slug: h245_8_05_partial_slopes; caption: At the same point, the two coordinate sections have different tangent gradients.; description: Left plot has x and z axes and shows z=x squared+4 with its tangent z=2x+3 at (1,5), labelled y=2 and gradient 2. Right plot has y and z axes and shows z=1+y squared with tangent z=4y-3 at (2,5), labelled x=1 and gradient 4. Both marked points correspond to the surface point (1,2,5).]
At the same point, the two coordinate sections have different tangent gradients.

Worked example 1: powers with a constant coefficient

Question. For $f(x,y)=x^3y^2+4x-y^3$ , find $f_x$ and $f_y$ .

Plan. Differentiate once with respect to each variable, in separate calculations, changing which variable is held fixed.

\begin{aligned} f_x&=y^2(3x^2)+4-0 && y\text{ is constant}\\ &=3x^2y^2+4\\ f_y&=x^3(2y)+0-3y^2 && x\text{ is constant}\\ &=2x^3y-3y^2 \end{aligned}

Why this step? In $f_x$ , the term $-y^3$ is a constant, so it disappears. In $f_y$ , $4x$ disappears instead.

Answer. $f_x=3x^2y^2+4$ , $f_y=2x^3y-3y^2$ .

Check. On the section $y=1$ , $f=x^3+4x-1$ , whose derivative is $3x^2+4$ , agreeing with $f_x(x,1)$ .

Worked example 2 changes only $4x$ to $4xy$ . That term now contributes to both derivatives.

Worked example 2: a term containing both variables

Question. For $g(x,y)=x^3y^2+4xy-y^3$ , find $\partial g/\partial x$ and $\partial g/\partial y$ .

Plan. Hold the other coordinate fixed in each calculation, including in the new product $4xy$ .

\begin{aligned} \frac{\partial g}{\partial x} &=3x^2y^2+4y && \text{differentiate in }x\\ \frac{\partial g}{\partial y} &=2x^3y+4x-3y^2 && \text{differentiate in }y \end{aligned}

Why this step? $4xy$ is $4y$ times $x$ in the first calculation, but $4x$ times $y$ in the second.

Answer. $g_x=3x^2y^2+4y$ , $g_y=2x^3y+4x-3y^2$ .

Check. At $(1,2)$ , these give $g_x=20$ and $g_y=-4$ . The sections $g(x,2)=4x^3+8x-8$ and $g(1,y)=y^2+4y-y^3$ have exactly those gradients at $x=1$ and $y=2$ .

Constants can be negative powers of the other variable too. For $z=x/y=xy^{-1}$ , where $y\ne0$ , the same rule gives $z_x=1/y$ and $z_y=-x/y^2$ .

Watch out

Wrong: $\partial(x^2y^3)/\partial x=2x\cdot3y^2$ . That differentiates both independent variables at once. Right: $\partial(x^2y^3)/\partial x=2xy^3$ , with $y^3$ unchanged.

The next check changes the coefficient of the mixed term and asks for the gradients at a specified point.

Product and chain rules

A partial derivative uses the familiar differentiation rules, with the other variable acting as a constant throughout.

For example, in differentiating $\sin(xy)$ with respect to $x$ , the inner function $xy$ has derivative $y$ , so the result is $y\cos(xy)$ . With respect to $y$ , the inner derivative is $x$ , giving $x\cos(xy)$ . The two chain-rule factors differ.

Chain rule for a function of two independent variables

If $u=u(x,y)$ and $F$ is differentiable, then

\frac{\partial}{\partial x}F(u)=F'(u)\frac{\partial u}{\partial x}, \qquad \frac{\partial}{\partial y}F(u)=F'(u)\frac{\partial u}{\partial y}.

Hold the other independent variable fixed each time.
Not in the formula booklet: learn it.

For $\ln(x+y)$ , both inner derivatives are $1$ , so both first partial derivatives are $1/(x+y)$ , where $x+y>0$ . For $e^{xy}$ , the inner derivatives are $y$ and $x$ . Write the relevant inner derivative before simplifying.

Worked example 3: two chain rules

Question. For $f(x,y)=e^{xy}+\ln(x+y)$ , where $x+y>0$ , find $f_x$ and $f_y$ .

Plan. Apply the chain rule to each term, separately for each independent variable.

\begin{aligned} f_x&=e^{xy}\,y+\frac{1}{x+y}\cdot1 && \partial(xy)/\partial x=y\\ f_y&=e^{xy}\,x+\frac{1}{x+y}\cdot1 && \partial(xy)/\partial y=x \end{aligned}

Why this step? The outer derivative of the exponential stays the same, but the inner derivative changes with the coordinate.

Answer. $f_x=ye^{xy}+1/(x+y)$ , $f_y=xe^{xy}+1/(x+y)$ .

Check. On $y=0$ , the original section is $1+\ln x$ , whose derivative is $1/x$ . The formula gives $f_x(x,0)=1/x$ too.

Worked example 4 adds just one factor, $x$ , in front of the exponential. The $x$ -derivative now needs a product rule, while the $y$ -derivative treats that factor as constant.

Worked example 4: product and chain together

Question. For $g(x,y)=xe^{xy}+\ln(x+y)$ , where $x+y>0$ , find $g_x$ and $g_y$ .

Plan. Use a product rule in $x$ , then differentiate with respect to $y$ holding the leading $x$ fixed.

\begin{aligned} g_x&=1\cdot e^{xy}+x(ye^{xy})+\frac1{x+y} && \text{product rule in }x\\ &=(1+xy)e^{xy}+\frac1{x+y} && \text{collect exponential terms}\\ g_y&=x(xe^{xy})+\frac1{x+y} && x\text{ is constant}\\ &=x^2e^{xy}+\frac1{x+y} \end{aligned}

Why this step? In $g_x$ , both factors $x$ and $e^{xy}$ depend on $x$ ; differentiating only the second factor would lose the term $e^{xy}$ .

Answer. $g_x=(1+xy)e^{xy}+1/(x+y)$ , $g_y=x^2e^{xy}+1/(x+y)$ .

Check. On $y=0$ , $g=x+\ln x$ , so $g_x(x,0)=1+1/x$ , agreeing with the formula.

The ordinary product rule is recalled knowledge; its partial-derivative version is not printed separately in the formula booklet.

Watch out

Wrong: $\partial e^{xy}/\partial x=e^{xy}$ . Right: it is $ye^{xy}$ . The missing factor is the derivative of the exponent with respect to the variable named in the denominator.

A power of the leading $x$ changes the product-rule term once more.

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8.05d - First and second partial derivatives

First partial derivatives

Product and chain rules

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