PARTIAL DERIVATIVES The rules that apply to functions of 1 variable also to functions of 2 or more variables. However, a function of 2 variables can change in response to a change in either of its variables. If one of its variables is fixed and the other changed infinitesimally, the ratio of the resultant infinitesimal change in the function to the change in that variable is the PARTIAL DERIVATIVE of the function with respect to that variable. z = f ( x , y) f x( x0 , y0) = f y( x0 , y0) = lim x → x0 lim y → y0 f ( x , y0) − f ( x0 , y0) x − x0 f ( x 0 , y ) − f ( x 0 , y 0) y − y0 Page 1 of 14 Equivalent Notations: f x( x0 , y0) = ∂ ∂x f ( x0 , y0) f y( x0 , y0) = ∂ ∂y f ( x 0 , y 0) Higher order partial derivatives can also be computed for multivariable functions just like single variable functions. Example # 1: Find the first and ALL of the second partial derivatives of the given function. 2 f ( x , y) = x ⋅ y 2 Here are the First Partial Derivatives. ∂ ∂x f = 2⋅ x⋅ y 2 ∂ ∂y 2 f = 2⋅ x ⋅ y Here are the Second Partial Derivatives. Page 2 of 14 ⎞ ∂2 2 = = 2 ⋅ y f f ⎜ ⎟ 2 ∂x ⎝ ∂x ⎠ ∂x ∂ ⎛∂ ⎞ ∂ ∂ f = 4⋅ x⋅ y ⎜ f⎟ = ∂y ⎝ ∂x ⎠ ∂y ∂x ∂ ⎛∂ ⎞ ∂ ∂ f = 4⋅ x⋅ y ⎜ f⎟ = ∂x ⎝ ∂y ⎠ ∂x ∂y ∂ ⎛∂ ⎞ ∂2 2 f = = 2 ⋅ x f ⎜ ⎟ 2 ∂y ⎝ ∂y ⎠ ∂y ∂ ⎛∂ Note that in this case that the Mixed Second Partial Derivatives are equal. They are always equal if they are both continuous functions. ∂ ∂ ∂x ∂y f = ∂ ∂ ∂y ∂x f Page 3 of 14 Definition: Let " z = f ( x , y) " be differentiable at the point: ( x0 , y0). Denote a new function, " dz = f x( x0 , y0) ⋅ dx + f y( x0 , y0) ⋅ dy " with dependent variable " dz " and independent variables: " dx " and " dy ". This "new" function is refered to as the Total Differential. Example # 2: Compute the total differential of the given function at the point: ( 1 , 2 , 4). 2 z = f ( x , y) = x ⋅ y 2 ⎛∂ ⎞ ⎛∂ ⎞ dz = ⎜ f ⎟ ⋅ dx + ⎜ f ⎟ ⋅ dy ⎝ ∂x ⎠ ⎝ ∂y ⎠ ∂ ∂x f = 2⋅ x⋅ y ( 2 ) 2 d 2 f = 2⋅ x ⋅ y dy ( 2 ) dz = 2 ⋅ x ⋅ y ⋅ dx + 2 ⋅ x ⋅ y ⋅ dy 2 2 dz = ⎡⎣ 2 ⋅ ( 1) ⋅ ( 2) ⎤⎦ ⋅ dx + ⎡⎣2 ⋅ ( 1) ⋅ ( 2)⎤⎦ ⋅ dy dz = ( 8) ⋅ dx + ( 4) ⋅ dy Page 4 of 14 Example # 3: Compute the total differential of the given function at the arbitrary point: ( x , y , f ( x , y) ). z = f ( x , y) = tan − 1( x ⋅ y) ⎛∂ ⎞ ⎛∂ ⎞ dz = ⎜ f ⎟ ⋅ dx + ⎜ f ⎟ ⋅ dy ⎝ ∂x ⎠ ⎝ ∂y ⎠ ∂ ∂x ∂ ∂y f = ⎛⎜ ⎞⋅ y ⎟ + ⋅ 1 x y ⎝ ⎠ 1 f = ⎛⎜ 1 ⎞⋅⎛ x ⎟⎜ + ⋅ 1 x y ⎝ ⎠ ⎝ 2⋅ ⎛ y ⎞ ⎟ y⎠ ⎞ x ⎡ ⎤ ⋅ dy ⋅ dx + ⎟ ⎢ ⎥ ( ) 1 + x ⋅ y ⋅ ⋅ 1 + x ⋅ y 2 y ⎝ ⎠ ⎣ ⎦ dz = ⎜ The problems that follow illustrate application of the total differential. Page 5 of 14 Example # 4: A point moves along the intersection of the given elliptic paraboloid and plane. At what rate is "z " changing with "x" when the point is at ( 1 , 1 , 4) ? Also, find the equation of the tangent line at that point. 2 z ( x , y) = x + 3 ⋅ y 2 y=1 Paraboloid & Plane z y x 2 z ( x , y) = x + 3 ⋅ y 2 y=1 The two surfaces intersect here. 2 z ( x , 1) = x + 3 Page 6 of 14 dz = 2 ⋅ x ⋅ dx dz = 2 ⋅ ( 1) ⋅ dx dy = 0 Tangent Line equations are thus these. x ( t ) = 1 + ( 1) ⋅ t y ( t ) = 1 + ( 0) ⋅ t z ( t ) = 4 + ( 2) ⋅ t ⎡ 1 + ( 1) ⋅ t ⎤ ⎯⎯→ ⎢ ⎥ LT ( t) = ⎢ 1 + ( 0) ⋅ t ⎥ ⎢ 4 + ( 2) ⋅ t ⎥ ⎣ ⎦ Those are the parametric form for the equation of the tangent line at the given point on the curve defined by the intersection of the elliptic paraboloid and the plane. Page 7 of 14 Tangent@ (1,1,4) z y x Example # 5: Find the equation of the tangent line at the point : ( 1, 1, 4 ) on the curve defined by the intersection of the given elliptic paraboloid and plane. 2 z1 ( x , y) = x + 3 ⋅ y 2 z2 ( x , y) = 2 ⋅ x + 2 ⋅ y Page 8 of 14 Parabaloid & Tilted Plane z x y dz1 = 2 ⋅ x ⋅ dx + 6 ⋅ y ⋅ dy 2 dz2 = 2 ⋅ dx + 2 ⋅ dy 2 x + 3⋅ y = 2⋅ x + 2⋅ y ⎛d ⎞ d 1 + 1⋅ y = 1⋅ x + 3⋅ y⋅ ⎜ y⎟ dx ⎝ dx ⎠ Page 9 of 14 ⎛d ⎞ ⎜ y ⎟ ⋅ ( 3 ⋅ y − 1) = 1 − x ⎝ dx ⎠ 1−x d y= 3⋅ y − 1 dx 1 − ( 1) d y= =0 3 ⋅ ( 1) − 1 dx dy = ( 0)dx dz1 = 2 ⋅ x ⋅ dx + 6 ⋅ y ⋅ dy dz1 = 2 ⋅ x ⋅ dx + 6 ⋅ y ⋅ ( 0) ⋅ dx dz1 = 2 ⋅ ( 1) ⋅ dx dz2 = 2 ⋅ dx + 6 ⋅ dy dz2 = 2 ⋅ dx + 6 ⋅ ( 0) ⋅ dx dz2 = 2 ⋅ dx dz1 = dz2 = dz = ( 2) ⋅ dx ⎡ 1 + ( 1) ⋅ t ⎤ ⎯⎯→ ⎢ ⎥ LT ( t) = ⎢ 1 + ( 0) ⋅ t ⎥ ⎢ 4 + ( 2) ⋅ t ⎥ ⎣ ⎦ Page 10 of 14 Tangent@ (1,1,4) z x y Example # 6: Find the equation of the tangent line at the point:" ( 1 , 0 , 1)" on the curve defined by the equation of the intersection of the given surfaces. Page 11 of 14 2 z1 ( x , y) = 2 − x − y 2 2 z2 ( x , y) = x + y 2 Curved Surfaces z y x 2 z1 ( x , y) = 2 − x − y 2 2 2 z2 ( x , y) = x + y 2 2 2−x −y = x +y −2 ⋅ x − 2 ⋅ y ⋅ d d y = 2⋅ x + y dx dx Page 12 of 14 ( 1 + 2 ⋅ y) ⋅ d y = −4 ⋅ x dx −4 ⋅ x d y= 1 + 2⋅ y dx Substitute the values of the "x" and "y" coordinates of the given point: ( 1 , 0 , 1). −4 ⋅ ( 1) d y= = −4 1 + 2 ⋅ ( 0) dx dy = −4 ⋅ dx dz1 = −2 ⋅ x ⋅ dx − 2 ⋅ y ⋅ dy dz2 = 2 ⋅ x + 1 ⋅ dy dx==−12⋅⋅ dx dz1 x ⋅ dx − 2 ⋅ y ⋅ ( −4 ⋅ dx) dz2 = 2 ⋅ x + 1 ⋅ ( −4 ⋅ dx) dz1 = ( −2 ⋅ x + 8 ⋅ y) ⋅ dx dz2 = ( 2 ⋅ x − 4) ⋅ dx Page 13 of 14 dz1 = [ −2 ⋅ ( 1) + 8 ⋅ ( 0) ] ⋅ dx dz2 = [ 2 ⋅ ( 1) − 4] ⋅ dx dz1 = dz2 = dz = ( −2) ⋅ dx ⎡ 1 + ( 1) ⋅ t ⎤ ⎯⎯→ ⎢ ⎥ LT ( t) = ⎢ 0 + ( −4) ⋅ t ⎥ ⎢ 1 + ( −2) ⋅ t ⎥ ⎣ ⎦ Tangent@(1,0,1) z y x Page 14 of 14