Finding the derivative when you can’t solve for y
Implicit Differentiation In many examples, especially the ones derived from differential equations, the variables involved are not linked to each other in an explicit way. Most of the time, they are linked through an implicit formula, like F ( x, y ) =0. Every y=f (x) is an explicit function because it is clear that the value of y is dependent on the value of x. On the other side, an implicit function is any 'function' where there doesn't appear to be any dependent variable, such as x^2+y^2=1. To put simply because the calculators were designed as such. Lets have some history. First Casio calculators made before 1978 or so had no operator precedence. They calculated everything left to right so 1+2.3 gave 9.
You may like to read Introduction to Derivatives and Derivative Rules first.
A function can be explicit or implicit:
Explicit: 'y = some function of x'. When we know x we can calculate y directly.
Implicit: 'some function of y and x equals something else'. Knowing x does not lead directly to y.
|Explicit Form||Implicit Form|
|y = ± √ (r2 − x2)||x2 + y2 = r2|
|In this form, y is expressed|
as a function of x.
|In this form, the function is |
expressed in terms of both y and x.
The graph of x2 + y2 = 32
Differentiate with respect to x:
ddx(x2) + ddx(y2) = ddx(r2)
Let's solve each term:
Which gives us:
2x + 2ydydx = 0
Collect all the dydx on one side
ydydx = −x
Solve for dydx:
dydx = −xy
Let's look more closely at how ddx(y2) becomes 2ydydx
The Chain Rule says:
dudx = dudydydx
Substitute in u = y2:
ddx(y2) = ddy(y2)dydx
ddx(y2) = 2ydydx
Another common notation is to use ’ to mean ddx
The Chain Rule can also be written using ’ notation:
f(g(x))’ = f’(g(x))g’(x)
g(x) is our function 'y', so:
f(y)’ = f’(y)y’
f(y) = y2, so f’(y) = 2y:
f(y)’ = 2yy’
or alternatively: f(y)’ = 2y dydx
Let's also find the derivative using the explicit form of the equation.
We get the same result this way!
You can try taking the derivative of the negative term yourself.
Yes, we used the Chain Rule again. Like this (note different letters, but same rule):
dydx = dydfdfdx
Substitute in f = (r2 − x2):
ddx(f½) = ddf(f½)ddx(r2 − x2)
ddx(f½) = ½(f−½) (−2x)
And substitute back f = (r2 − x2):
ddx(r2 − x2)½ = ½((r2 − x2)−½) (−2x)
And we simplified from there.
OK, so why find the derivative y’ = −x/y ?
Well, for example, we can find the slope of a tangent line.
No problem, just substitute it into our equation:
dydx = −x/y
dydx = −3/4
And for bonus, the equation for the tangent line is:
y = −3/4 x + 25/4
Sometimes the implicit way works where the explicit way is hard or impossible.
How do we solve for y? We don't have to!
(the middle term is explained below)
And we get:
|dydx =||9y2 − 20x3|
|3(5y2 − 6xy)|
For the middle term we used the Product Rule: (fg)’ = f g’ + f’ g
Because (y2)’ = 2ydydx(we worked that out in a previous example)
Oh, and dxdx = 1, in other words x’ = 1
Implicit differentiation can help us solve inverse functions.
The general pattern is:
As a final step we can try to simplify more by substituting the original equation.
An example will help:
We can also go one step further using the Pythagorean identity:
sin2 y + cos2 y = 1
cos y = √(1 − sin2 y )
And, because sin(y) = x (from above!), we get:
cos y = √(1 − x2)
Which leads to:
dydx= 1√(1 − x2)
Note: this is the same answer we get using the Power Rule: