Section 15.4 Partial Derivatives Chapter 15 Section 15.4 Partial Derivatives

Partial Derivatives The partial derivative of the variable 𝑧=𝑓 𝑥,𝑦 with respect to either the variables x or y is given by each of the limits to the right. 𝜕𝑧 𝜕𝑥 = 𝑓 𝑥 𝑥,𝑦 = lim ℎ→0 𝑓 𝑥+ℎ,𝑦 −𝑓 𝑥,𝑦 ℎ 𝜕𝑧 𝜕𝑦 = 𝑓 𝑦 𝑥,𝑦 = lim ℎ→0 𝑓 𝑥,𝑦+ℎ −𝑓 𝑥,𝑦 ℎ Computing Partial Derivatives To make use of the differentiation rules learned in earlier calculus courses we realize that these derivatives only deal with 1 “variable” at a time. To calculate the derivative of the dependent variable (z for 𝑧=𝑓 𝑥,𝑦 ) with respect to an independent variable (x or y for 𝑧=𝑓 𝑥,𝑦 ) view the independent variable as the “only variable” and all other variables as constants. For 𝑧=𝑓 𝑥,𝑦 To find 𝜕𝑧 𝜕𝑥 take the derivative of z viewing x as the variable and y and any expression dependent on y as a constant. To find 𝜕𝑧 𝜕𝑦 take the derivative of z viewing y as the variable and x and any expression dependent on x as a constant. Example Find 𝜕𝑧 𝜕𝑥 and 𝜕𝑧 𝑑𝑦 for the function 𝑧=𝑓 𝑥,𝑦 = 𝑥 5 cos 𝑦 + ln 𝑦 −𝑥 𝑒 𝑥 𝑦 2 𝜕𝑧 𝜕𝑥 = 𝜕 𝜕𝑥 𝑥 5 cos 𝑦 + 𝜕 𝜕𝑥 ln 𝑦 − 𝜕 𝜕𝑥 𝑥 𝑒 𝑥 𝑦 2 =5 𝑥 4 cos 𝑦 +0− 𝑒 𝑥 𝑦 2 +𝑥 𝑦 2 𝑒 𝑥 𝑦 2 =5 𝑥 4 cos 𝑦 − 𝑒 𝑥 𝑦 2 −𝑥 𝑦 2 𝑒 𝑥 𝑦 2 𝜕𝑧 𝜕𝑦 = 𝜕 𝜕𝑦 𝑥 5 cos 𝑦 + 𝜕 𝜕𝑦 ln 𝑦 − 𝜕 𝜕𝑦 𝑥 𝑒 𝑥 𝑦 2 =− 𝑥 5 sin 𝑦 + 1 𝑦 −𝑥 𝑒 𝑥 𝑦 2 2𝑥𝑦 =− 𝑥 5 sin 𝑦 + 1 𝑦 −2 𝑥 2 𝑦 𝑒 𝑥 𝑦 2

Partial Derivatives with more Variables If there are more independent variables for the function than x and y view all the independent variables that you are not taking the derivative with respect to as constants. For 𝑤=𝑓 𝑥,𝑦,𝑧 Find 𝜕𝑤 𝜕𝑧 by taking the derivative of w viewing z as the variable and x and y and all expressions depending only on x or y as constants Example Find 𝜕𝑤 𝜕𝑧 for the function 𝑤=𝑓 𝑥,𝑦,𝑧 = 𝑒 −2𝑥 sec 𝑦 −𝑦 sinh 𝑧 + 𝑥 2 arctan 𝑧 𝜕𝑤 𝜕𝑧 = 𝜕 𝜕𝑧 𝑒 −2𝑥 sec 𝑦 − 𝜕 𝜕𝑧 𝑦 sinh 𝑧 + 𝜕 𝜕𝑧 𝑥 2 arctan 𝑧 =−𝑦 cosh 𝑧 + 𝑥 2 1+ 𝑧 2 Implicit Partial Derivatives Some expression are very complicated or even impossible to solve for the dependent variable, but it still might be useful to know the partial derivative. To do this we can take the partial derivative implicitly. Steps for implicit derivatives Take derivative of both sides of equation keeping in mind independent variable, multiplying by derivative where appropriate. Solve for derivative Example For the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 =16 find 𝜕𝑧 𝜕𝑥 . Implicitly, keeping in mind 𝑧=𝑓 𝑥,𝑦 . 𝜕 𝜕𝑥 𝑥 2 + 𝜕 𝜕𝑥 𝑦 2 + 𝜕 𝜕𝑥 𝑧 2 = 𝜕 𝜕𝑥 16 2𝑥+0+2𝑧 𝜕𝑧 𝜕𝑥 =0 𝜕𝑧 𝜕𝑥 = −2𝑥 2𝑧 = −𝑥 𝑧 Solve for z : 𝑧=± 16− 𝑥 2 − 𝑦 2 𝜕𝑧 𝜕𝑥 = −2𝑥 ±2 16− 𝑥 2 − 𝑦 2 = −𝑥 𝑧

Geometric View of Derivative Remember for a function of 1 variable the derivative at a point 𝑥 0 represents the slope of the tangent line or if you go 1 unit in the x direction you go up or down on the tangent line by the amount of the derivative 𝑑𝑦 𝑑𝑥 𝑥= 𝑥 0 . y 𝑑𝑦 𝑑𝑥 𝑥= 𝑥 0 1 x 𝑥 0 Geometric View of Partial Derivative For a surface 𝑧=𝑓 𝑥,𝑦 if we fix a value for x or y say 𝑥 0 and 𝑦 0 this makes a curve where the surface intersects 𝑥= 𝑥 0 and 𝑦= 𝑦 0 . These curves have tangent lines at the point 𝑥 0 , 𝑦 0 . The partial derivative tells you if you go 1 unit forward in the x direction or 1 unit right in the y direction how much you go up or down in the z direction. x y z 𝑥 0 , 𝑦 0 1 𝜕𝑧 𝜕𝑥 𝑥= 𝑥 0 𝑦= 𝑦 0 𝑥 0 𝑦 0 x y z 𝑥 0 , 𝑦 0 1 𝜕𝑧 𝜕𝑦 𝑥= 𝑥 0 𝑦= 𝑦 0 𝑥 0 𝑦 0 Direction Vector of Tangent line: 1,0, 𝜕𝑧 𝜕𝑥 Direction Vector of Tangent line: 0,1, 𝜕𝑧 𝜕𝑦