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