1
Implicit Differentiation
Often we have functions that are explicitly given as 𝑦 in terms of 𝑥 like
𝑦 = √𝑥 2 + 1 or 𝑦 = 𝑥(𝑠𝑖𝑛𝑥). But sometimes the relation between 𝑦 and 𝑥 is
given implicitly by an equation like 𝑥 2 + 𝑦 2 = 1 or 𝑦 2 = 𝑥 . Sometimes it’s
not even possible to solve for 𝒚 in terms of 𝒙. The points that satisfy these
equations form a curve in the plane. The question is, how do we find the slope of
𝑑𝑦
the tangent line (if it exists) to a point on one of these curves? To find in this
𝑑𝑥
case (which is the slope of the tangent line at each point (𝑥, 𝑦) on the curve) we
need to do it through implicit differentiation.
𝑑𝑦
Ex. Find for the circle given by 𝑥 2 + 𝑦 2 = 25. Find an equation for the
𝑑𝑥
tangent line at (−3,4). Where is the tangent line horizontal?
To do this we treat 𝑦 as a function of 𝑥 , ie 𝑦(𝑥), and differentiate the
equation using the chain rule. We differentiate any pure expressions in 𝑥 using
our differentiation rules.
𝑑 𝑑
𝑑𝑥
(𝑥2 + 𝑦2 ) = 𝑑𝑥 (25)
𝑑 𝑑 𝑑
(𝑥 2 ) + (𝑦 2 ) = (25)
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑
2𝑥 + (𝑦 2 ) = 0.
𝑑𝑥
𝑑
To calculate (𝑦 2 ) treat 𝑦 as 𝑦(𝑥) and use the chain rule.
𝑑𝑥
𝑑 𝑑 𝑑𝑦
(𝑦 2 ) = (𝑦(𝑥)2 ) = 2(𝑦(𝑥 ))𝑦 ′ (𝑥 ) = 2𝑦(𝑥 ) .
𝑑𝑥 𝑑𝑥 𝑑𝑥
, 2
𝑑𝑦
So we have: 2𝑥 + 2𝑦 = 0.
𝑑𝑥
𝑑𝑦 𝑑𝑦
Now solve this equation for ∶ 2𝑦 = −2𝑥
𝑑𝑥 𝑑𝑥
𝑑𝑦 𝑥
= −𝑦 .
𝑑𝑥
3
So the slope of the tangent line at (−3,4) is ,
4
3
An equation of the tangent line at (−3,4) is: 𝑦 − 4 = (𝑥 + 3).
4
𝑑𝑦
The tangent line is horizontal when = 0.
𝑑𝑥
𝑑𝑦 −𝑥
= = 0 when 𝑥 = 0.
𝑑𝑥 𝑦
Solving for 𝑦, we get: 02 + 𝑦 2 = 25 means 𝑦 = −5, 5.
So horizontal tangent lines at (0, −5), (0,5).
So when differentiating an equation with 𝒙’s and 𝒚’s in it, differentiate the
terms with just 𝒙 in them as we have before. Every time you differentiate a
𝒅𝒚
term with a 𝒚 in it, you will get a coming from the chain rule.
𝒅𝒙
Implicit Differentiation
Often we have functions that are explicitly given as 𝑦 in terms of 𝑥 like
𝑦 = √𝑥 2 + 1 or 𝑦 = 𝑥(𝑠𝑖𝑛𝑥). But sometimes the relation between 𝑦 and 𝑥 is
given implicitly by an equation like 𝑥 2 + 𝑦 2 = 1 or 𝑦 2 = 𝑥 . Sometimes it’s
not even possible to solve for 𝒚 in terms of 𝒙. The points that satisfy these
equations form a curve in the plane. The question is, how do we find the slope of
𝑑𝑦
the tangent line (if it exists) to a point on one of these curves? To find in this
𝑑𝑥
case (which is the slope of the tangent line at each point (𝑥, 𝑦) on the curve) we
need to do it through implicit differentiation.
𝑑𝑦
Ex. Find for the circle given by 𝑥 2 + 𝑦 2 = 25. Find an equation for the
𝑑𝑥
tangent line at (−3,4). Where is the tangent line horizontal?
To do this we treat 𝑦 as a function of 𝑥 , ie 𝑦(𝑥), and differentiate the
equation using the chain rule. We differentiate any pure expressions in 𝑥 using
our differentiation rules.
𝑑 𝑑
𝑑𝑥
(𝑥2 + 𝑦2 ) = 𝑑𝑥 (25)
𝑑 𝑑 𝑑
(𝑥 2 ) + (𝑦 2 ) = (25)
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑
2𝑥 + (𝑦 2 ) = 0.
𝑑𝑥
𝑑
To calculate (𝑦 2 ) treat 𝑦 as 𝑦(𝑥) and use the chain rule.
𝑑𝑥
𝑑 𝑑 𝑑𝑦
(𝑦 2 ) = (𝑦(𝑥)2 ) = 2(𝑦(𝑥 ))𝑦 ′ (𝑥 ) = 2𝑦(𝑥 ) .
𝑑𝑥 𝑑𝑥 𝑑𝑥
, 2
𝑑𝑦
So we have: 2𝑥 + 2𝑦 = 0.
𝑑𝑥
𝑑𝑦 𝑑𝑦
Now solve this equation for ∶ 2𝑦 = −2𝑥
𝑑𝑥 𝑑𝑥
𝑑𝑦 𝑥
= −𝑦 .
𝑑𝑥
3
So the slope of the tangent line at (−3,4) is ,
4
3
An equation of the tangent line at (−3,4) is: 𝑦 − 4 = (𝑥 + 3).
4
𝑑𝑦
The tangent line is horizontal when = 0.
𝑑𝑥
𝑑𝑦 −𝑥
= = 0 when 𝑥 = 0.
𝑑𝑥 𝑦
Solving for 𝑦, we get: 02 + 𝑦 2 = 25 means 𝑦 = −5, 5.
So horizontal tangent lines at (0, −5), (0,5).
So when differentiating an equation with 𝒙’s and 𝒚’s in it, differentiate the
terms with just 𝒙 in them as we have before. Every time you differentiate a
𝒅𝒚
term with a 𝒚 in it, you will get a coming from the chain rule.
𝒅𝒙
