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Wize University Calculus 3 Textbook > Partial Derivatives

Partial Derivatives

Tangent, Normal, Binormal Vectors - Applications of Vector FunctionsPrevious Section
Chain Rules for Functions with Several Variables & Implicit DifferentiationNext Section
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Partial Derivatives


To perform the partial derivative of f(x,y)f(x,y) with respect to variable xx, we assume that the other variable, yy, is constant (i.e. we treat it like a constant when we differentiate).

In this way, partial derivatives will be similar to the derivative of functions with only one variable.

Theorem

If a given function ff has two variables, the partial derivatives of this function with respect to xx and yy are usually denoted as:

fx(x,y)=fx=∂f∂xfy(x,y)=fy=∂f∂y\begin{array}{rcl} f_x(x,y)&=&f_x&=&\dfrac{\partial{f}}{\partial{x}}\\\\ f_y(x,y)&=&f_y&=&\dfrac{\partial{f}}{\partial{y}} \end{array}
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Higher Order Partial Derivatives & Differentials


Second-order partial derivatives

(fx)x=fxx=∂x(∂f∂x)=∂2f∂x2(fx)y=fxy=∂y(∂f∂x)=∂2f∂y∂x(fy)x=fyx=∂x(∂f∂y)=∂2f∂x∂y(fy)y=fyy=∂y(∂f∂y)=∂2f∂y2\begin{array}{rcl} (f_x)_x&=&f_{xx}&=&\dfrac{\partial}{x}\Big(\dfrac{\partial{f}}{\partial{x}}\Big)&=&\dfrac{\partial^2{f}}{\partial{x^2}}\\\\ (f_x)_y&=&f_{xy}&=&\dfrac{\partial}{y}\Big(\dfrac{\partial{f}}{\partial{x}}\Big)&=&\dfrac{\partial^2{f}}{\partial{y}\partial{x}}\\\\ (f_y)_x&=&f_{yx}&=&\dfrac{\partial}{x}\Big(\dfrac{\partial{f}}{\partial{y}}\Big)&=&\dfrac{\partial^2{f}}{\partial{x}\partial{y}}\\\\ (f_y)_y&=&f_{yy}&=&\dfrac{\partial}{y}\Big(\dfrac{\partial{f}}{\partial{y}}\Big)&=&\dfrac{\partial^2{f}}{\partial{y^2}}\\\\ \end{array}

Clairaut's Theorem

Suppose that ff is defined on a disk D that contains (a,b)(a,b).

If the functions fxy and fyx f_{xy}~\text{and}~f_{yx}~ are continuous on this disk, then:
fxy(a,b)=fyx(a,b)\boxed{f_{xy}(a,b)=f_{yx}(a,b)}

Wize Concept
Clairaut's Theorem extends to any functions and their mixed partial derivatives al long as continuity rule applies

Differentials


  • Given z=f(x, y), z=f(x,~y),~ the differential ∂z\partial{z} or ∂f\partial{f} is:
∂z=fx∂x+fy∂y       or      ∂f=fx∂x+fy∂y\partial{z}=f_x\partial{x}+f_y\partial{y}~~~~~~~{\color{magenta}\text{\scriptsize{or}}}~~~~~~\partial{f}=f_x\partial{x}+f_y\partial{y}

  • Given f(x, y, z)=wf(x,~y,~z)=w, then:
∂w=fx∂x+fy∂y+fz∂z\partial{w}=f_x\partial{x}+f_y\partial{y}+f_z\partial{z}

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Example

Calculate ∂f∂x\dfrac{\partial f}{\partial x}, ∂f∂y\dfrac{\partial f}{\partial y}, ∂2f∂x2\dfrac{\partial^2f}{\partial x^2}, ∂2f∂y2\dfrac{\partial^2f}{\partial y^2} for f(x,y)=excos⁡(y)f(x,y)=e^x\cos(y).

f(x,y)=excos⁡(y)f(x,y)=e^x\cos(y)

∂f∂x=fx=excos⁡(y)∂f∂y=fy=−exsin⁡(y)\begin{array}{rcl} \dfrac{\partial f}{\partial x}&=&f_x\\\\ &=&e^x\cos(y)\\\\ \qquad\dfrac{\partial f}{\partial y}&=&f_y\\\\ &=&{-e^x\sin(y)} \end{array} ∂2f∂x2=fxx=excos⁡(y)∂2f∂y2=fyy=−excos⁡(y)\begin{array}{rcl} \dfrac{\partial^2f}{\partial{x^2}}&=&f_{xx}&=&e^x\cos(y)\\\\ \dfrac{\partial^2f}{\partial y^2}&=&f_{yy}&=&-e^x\cos(y) \end{array}

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Example

Determine if z=x2y3 z=\frac{x^2}{y^3}~ is increasing or decreasing at (2,5)(2,5) if we hold y fixed.

Let z=f(x, y)=x2y3. z=f(x,~y)=\frac{x^2}{y^3}.~

If we hold y fixed, then we treat 'y' like a constant and find fxf_x:
fx(x, y)=2xy3fx(2, 5)=2×253fx(2, 5)=4125>0\begin{array}{rcl} f_x(x,~y)&=&\dfrac{2x}{y^3}\\\\ f_x(2,~5)&=&\dfrac{2\times2}{5^3}\\\\ f_x(2,~5)&=&\dfrac{4}{125}>0 \end{array}

Our function is increasing at (2,5)(2,5).

Practice

Find the first partial derivatives of the function z=4x3exy+4xz=4x^{3}e^{xy}+\frac{4}{x}

Practice

For what values of k does r(x,y)=e2kxcos⁡(4y) r(x, y) = e^{2kx}\cos(4y)~ satisfy 2rx=ryy?2r_x=r_{yy}?