MATH 12002 - CALCULUS I § 2.6: Implicit Differentiation Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 7
Implicit Differentiation A relation between x and y represents a graph in the xy -plane; i.e., the set of points ( a , b ) whose coordinates x = a , y = b satisfy the relation. If the relation is not a function, we will not be able to write y as a single function of x . However, if the graph is a smooth curve, it will still have tangent lines and we should be able to find the slopes of the tangent lines, even though y is not of the form y = f ( x ) and the slope cannot be written as f ′ ( a ). We will see how to find this slope (the derivative of y with respect to x ) implicitly , but first let’s consider an example where we can find the slope explicitly . D.L. White (Kent State University) 2 / 7
Implicit Differentiation Example Find the slope of the line tangent to the ellipse 9 x 2 + 4 y 2 = 36 at the √ point (1 , − 3 3 2 ) . Solution 1 We can solve for y in terms of x: since 9 x 2 + 4 y 2 = 36 , we have 4 y 2 36 − 9 x 2 = y 2 1 4 (36 − 9 x 2 ) = � ± 1 36 − 9 x 2 . y = 2 √ 36 − 9 x 2 represents the top half of the ellipse (y � 0 ) and Here, y = 1 √ 2 36 − 9 x 2 represents the bottom half (y � 0 ). y = − 1 [Continued → ] 2 D.L. White (Kent State University) 3 / 7
Implicit Differentiation Solution 1 [continued] √ Since our point, (1 , − 3 3 2 ) , is on the bottom half of the ellipse, the slope √ is f ′ (1) , where f ( x ) = − 1 36 − 9 x 2 . We have 2 2 (36 − 9 x 2 ) − 1 / 2 · ( − 18 x ) − 1 2 · 1 f ′ ( x ) = 9 x = √ 36 − 9 x 2 , 2 and so the slope is √ 9(1) 9 9 3 · 3 3 f ′ (1) = = 2 √ 36 − 9 = √ = √ = � 2 . 36 − 9(1 2 ) 2 27 2 · 3 3 2 D.L. White (Kent State University) 4 / 7
Implicit Differentiation Solution 2 We now find the slope implicitly. We know that the top and bottom halves of the ellipse are functions. Instead of solving for the function explicitly, let’s just say y = f ( x ) , so that 9 x 2 + 4[ f ( x )] 2 = 36 . Now take derivatives of both sides of the equation with respect to x: 18 x + 8[ f ( x )] 1 · f ′ ( x ) = 0 . Since “f ( x ) = y” this means “f ′ ( x ) = dy dx ” and 18 x + 8 y · dy dx = 0 , √ and we get dy dx = − 9 x 4 y . The slope is then dy dx evaluated at (1 , − 3 3 2 ) , i.e., √ � dy 9(1) 3 · 3 3 � = − = √ = √ � 2 . dx 4( − 3 3 2 · 3 3 √ 2 ) � (1 , − 3 3 2 ) D.L. White (Kent State University) 5 / 7
Examples 1 Let x 2 + x 3 y 5 + y − 3 = 4 and find y ′ = dy dx , the derivative of y with respect to x . Instead of writing y as f ( x ), we will just treat y as some function of x , so that y 5 , for example, is a composite function with inside function y . First, take the derivative with respect to x of both sides of the equation x 2 + x 3 y 5 + y − 3 = 4, to obtain 2 x + [3 x 2 y 5 + x 3 · 5 y 4 y ′ ] − 3 y − 4 y ′ = 0 . Now solve for y ′ in terms of x and y : − 5 x 3 y 4 y ′ + 3 y − 4 y ′ 2 x + 3 x 2 y 5 = ( − 5 x 3 y 4 + 3 y − 4 ) y ′ 2 x + 3 x 2 y 5 = 2 x + 3 x 2 y 5 y ′ = − 5 x 3 y 4 + 3 y − 4 . D.L. White (Kent State University) 6 / 7
Examples 2 Let 2 y 3 = tan( x 2 + y 2 ) and find y ′ = dy dx , the derivative of y with respect to x . First, take the derivative with respect to x of both sides of the equation 2 y 3 = tan( x 2 + y 2 ) to obtain 6 y 2 y ′ = [sec 2 ( x 2 + y 2 )](2 x + 2 yy ′ ) . Now solve for y ′ in terms of x and y : 6 y 2 y ′ = [sec 2 ( x 2 + y 2 )] · 2 x + [sec 2 ( x 2 + y 2 )] · 2 yy ′ 6 y 2 y ′ − [sec 2 ( x 2 + y 2 )] · 2 yy ′ = [sec 2 ( x 2 + y 2 )] · 2 x 6 y 2 − [sec 2 ( x 2 + y 2 )] · 2 y y ′ = [sec 2 ( x 2 + y 2 )] · 2 x � � [sec 2 ( x 2 + y 2 )] · 2 x y ′ = 6 y 2 − [sec 2 ( x 2 + y 2 )] · 2 y . D.L. White (Kent State University) 7 / 7
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