Implicit Differentiation Department of Mathematics University of Leicester www.le.ac.uk What is it? • The normal function we work with is a function of the form: y f (x) • This is called an Explicit function. Example: y 2x 3 What is it? • An Implicit function is a function of the form: f ( x, y ) 0 • Example: y x y x 0 3 4 3 What is it? • This also includes any function that can be rearranged to this form: Example: y x y x 2y 3 4 3 • Can be rearranged to: y x y x 2y 0 3 4 3 • Which is an implicit function. What is it? • These functions are used for functions that would be very complicated to rearrange to the form y=f(x). • And some that would be possible to rearrange are far to complicated to differentiate. Implicit differentiation • To differentiate a function of y with respect to x, we use the chain rule. • If we have: z z ( y ), and y y (x) • Then using the chain rule we can see that: dz dz dy dx dy dx Implicit differentiation • From this we can take the fact that for a function of y: d d dy ( f ( x)) ( f ( y )) dx dy dx Implicit differentiation: example 1 • We can now use this to solve implicit differentials • Example: y x y x 2y 3 4 3 Implicit differentiation: example 1 • Then differentiate all the terms with respect to x: d d d d 3 d 3 4 (y ) ( x) ( y ) (x ) (2 y ) dx dx dx dx dx Implicit differentiation: example 1 • Terms that are functions of x are easy to differentiate, but functions of y, you need to use the chain rule. Implicit differentiation: example 1 • Therefore: d 3 d dy 3 2 dy (y ) (y ) 3y dx dy dx dx d d dy 4 4 3 dy (y ) (y ) 4y dx dy dx dx • And: d d dy dy (2 y ) (2 y ) 2 dx dy dx dx Implicit differentiation: example 1 • Then we can apply this to the original function: d d d d 3 d 3 4 (y ) ( x) ( y ) (x ) (2 y ) dx dx dx dx dx • Equals: dy dy 3 dy 2 3y 1 4y 3x 2 dx dx dx 2 Implicit differentiation: example 1 dy • Next, we can rearrange to collect the terms: dx dy dy 3 dy 2 3y 1 4y 3x 2 dx dx dx 2 • Equals: dy 3x 1 (4 y 3 y 2) dx 2 3 2 Implicit differentiation: example 1 • Finally: dy 3x 1 (4 y 3 y 2) dx 2 3 2 • Equals: dy 3x 1 3 2 dx 4 y 3 y 2 2 Implicit differentiation: example 2 • Example: cos y 2 x sin y 0 2 • Differentiate with respect to x d d cos y 4 x sin y 0 dx dx Implicit differentiation: example 2 • Then we can see, using the chain rule: d d dy dy cos y (cos y ) sin y dx dy dx dx • And: d d dy dy sin y (sin y ) cos y dx dy dx dx Implicit differentiation: example 2 • We can then apply these to the function: d d cos y 4 x sin y 0 dx dx • Equals: dy dy sin y 4 x cos y 0 dx dx Implicit differentiation: example 2 • Now collect the dy terms: dx dy 4 x (sin y cos y ) dx • Which then equals: dy 4x dx sin y cos y Implicit differentiation: example 3 • Example: x x y y 2 2 2 3 2 • Differentiate with respect to x d 2 d 2 3 d 2 d x x y y 2 dx dx dx dx Implicit differentiation: example 3 • To solve: d 2 3 x y dx • We also need to use the product rule. d 2 3 2 d 3 3 d x y x (y ) y (x2 ) dx dx dx dy 3 x .3 y y .2 x dx 2 2 Implicit differentiation: example 3 • We can then apply these to the function: d 2 d 2 3 d 2 d x x y y 2 dx dx dx dx • Equals: dy dy 3 2 x x .3 y y .2 x 2 y 0 dx dx 2 2 Implicit differentiation: example 3 • Now collect the dy terms: dx dy (2 y x 2 x) (2 y 3x y ) dx 3 2 • Which then equals: dy 2 y x 2x 2 2 dx 2 y 3x y 3 2 Implicit differentiation: example 4 • Example: x y arcsin( y ) 2 x 2 2 • Differentiate with respect to x d 2 d d 2 x arcsin( y ) 2x dx dx dx Implicit differentiation: example 4 • To solve: d 2 arcsin( y ) dx • We also need to use the chain rule, and the fact that: d 1 arcsin( x) 2 dx 1 x Implicit differentiation: example 4 • This is because: y arcsin( x) , so x sin( y ) • Therefore: dx 2 2 cos y 1 sin y 1 x dy • As x=sin(y), Therefore: dy d 1 arcsin( x) dx dx 1 x2 Implicit differentiation: example 4 • Using this, we can apply it to our equation : d 2 arcsin( y ) dx 1 dy 1 dy 2 y. 2 y. 2 2 4 dx dx 1 (y ) 1 y Implicit differentiation: example 4 • We can then apply these to the function: d 2 d d 2 x arcsin( y ) 2x dx dx dx • Equals: 2y dy 2x 2 4 dx 1 y Implicit differentiation: example 4 • Now collect the dy terms: dx 2y dy 2 2x 4 dx 1 y • Which then equals: dy (2 2 x)( 1 y ) dx 2y 4 Conclusion • Explicit differentials are of the form: y f (x) • Implicit differentials are of the form: f ( x, y ) 0 Conclusion • To differentiate a function of y with respect to x we can use the chain rule: d d dy ( f ( x)) ( f ( y )) dx dy dx