Subject: Calculus

Implicit Differentiation


In the previous examples, the functions presented are all explicit. However, when a function can not be defined explicitly, then we have to develop some technique in getting the derivative of that function. Example, a function f(x)=y=3x+\cos{x} is an explicit function. Notice that the right-hand side of the function contains only the variable x. So, we can easily differentiate such function using direct differentiation. But when a function look like these, x^{2}y+x\cos{xy}=5, then we really had a hard time converting these into an explicit function. Thus, we can't directly differentiate the function. Thus, we need to know about implicit differentiation.
Basically, implicit differentiation is just a process of differentiating an implicit function. The theorems and rules discussed above are still applicable in this process.

So how do we differentiate an implicit function?

Look at the following illustration.

Illustration

Consider the equation (x+y)^{2}-(x-y)^{2}=x^{4}+y^{4} +3. Isn't it hard to express that equation in terms of the variable x or y alone? Thus, we will do implicit differentiation in finding the derivative of that equation. What we will do here is to differentiate term by term to get

2(x+y)(1+y')-2(x-y)(1-y')=4x^{3}+4y^{3}y'+0
2x+2y+(2x+2y)y'-2x+2y+(2x-2y)y'=4x^{3}+4y^{3}y'
(4x-4y^{3})y'=4x^{3}+4y^{3}y'.

We then solve for y' which is given by

y'=\frac{4x^{3}-4y}{4x-4y^{3}}=\frac{x^{3}-y}{x-y^{3}}

Note that expressing the derivative in this form makes it easy to transform to other derivative notations such as \frac{dy}{dx}.

Example #1

Let y be a differentiable function of the variable x, then let us find its derivative.

x^2y=x+y

Doing the same thing above, we get

2xy + x^2y'=1+y'

Solving for y',

y'=\frac{2xy-1}{1-x^2}

Example #2

Let y be a differentiable function of the variable x, then let us find its derivative.

\sin{xy}=1

Doing the same thing above, we get

\cos{xy}\cdot D[xy]=0

\cos{xy}\cdot [y+xy']=0

Solving for y',

y'=-\frac{y}{x}

Example #3

Let y be a differentiable function of the variable x, then let us find its derivative.

5x^{3}y-7xy^{2}=9+7y

Doing the same thing above, we get

5(x^{3}y'+y \cdot 3x^{2})-7[x(2yy')+y^{2}]=0+7y'
5x^{3}y'+15x^{2}y-14xyy'-7y^{2}=7y

We simplify the above equation and get

(5x^{3}-14xy-7)y'=7y^{2}-15x^{2}y.

Solving for y',

y'=\frac{7y^{2}-15x^{2}y}{5x^{3}-14xy-7}

Example #4

Let y be a differentiable function of the variable x, then let us find its derivative.

y=\tan(x+y)

Using our theorems on differentiation of trigonometric functions, we have

y'=\sec^{2}(x+y)D_{x}(x+y)
 =(1+y')\sec^{2}(x+y)
 =\sec^{2}(x+y)+y'\sec^{2}(x+y)

Rearranging, we get

(1-\sec{2}(x+y))y'=\sec^{2}(x+y)

Therefore,

y'=\frac{\sec^{2}(x+y)}{1-\sec{2}(x+y)}

Example #5

Let y be a differentiable function of the variable x, then let us find its derivative.

\cot{xy}=-xy

Again, we differentiate term by term and get

(-\csc^{2}{xy})D_{x}(xy)+(xy'+y)=0
(-\csc^{2}{xy})(xy'+y)+(xy'+y)=0
-xy'\csc^{2}{xy}-y\csc^{2}{xy}+xy'+y=0
y'(x)(\csc^{2}{xy}-1)=y(1-\csc^{2}{xy})
y'=\frac{y}{x}\cdot \frac{1-\csc^{2}{xy}}{(\csc^{2}{xy}-1)}
y'=-\frac{y}{x}\cdot \frac{\csc^{2}{xy}-1}{(\csc^{2}{xy}-1)}=-\frac{x}{y}.

Therefore,

y'= -\frac{x}{y}.

NEXT TOPIC: Higher Order Derivatives