# Implicit Differentiation Examples

If you haven’t already read about implicit differentiation, you can read more about it here. Once you check that out, we’ll get into a few more examples below.

$$\mathbf{1. \ \ ycos(x) = x^2 + y^2}$$ | Solution

$$\mathbf{2. \ \ xy=x-y}$$ | Solution

$$\mathbf{3. \ \ x^2-4xy+y^2=4}$$ | Solution

$$\mathbf{4. \ \ \sqrt{x+y}=x^4+y^4}$$ | Solution

$$\mathbf{5. \ \ e^{x^2y}=x+y}$$ | Solution

For each of the above equations, we want to find dy/dx by implicit differentiation. In general a problem like this is going to follow the same general outline. Although, this outline won’t apply to every problem where you need to find dy/dx, this is the most common, and generally a good place to start. Start with these steps, and if they don’t get you any closer to finding dy/dx, you can try something else. Here are the steps:

• Take the derivative of both sides of the equation with respect to x.
• Separate all of the dy/dx terms from the non-dy/dx terms.
• Factor out the dy/dx.
• Isolate dy/dx.

Some of these examples will be using product rule and chain rule to find dy/dx.

## Solutions

$$\mathbf{1. \ \ ycos(x) = x^2 + y^2}$$

$$ycos(x)=x^2+y^2$$ $$\frac{d}{dx} \big[ ycos(x) \big] = \frac{d}{dx} \big[ x^2 + y^2 \big]$$ $$\frac{dy}{dx}cos(x) + y \big( -sin(x) \big) = 2x + 2y \frac{dy}{dx}$$ $$\frac{dy}{dx}cos(x) – y sin(x) = 2x + 2y \frac{dy}{dx}$$ $$\frac{dy}{dx}cos(x) -2y \frac{dy}{dx} = 2x + ysin(x)$$ $$\frac{dy}{dx} \big[ cos(x) -2y \big] = 2x + ysin(x)$$ $$\frac{dy}{dx} = \frac{2x + ysin(x)}{cos(x) -2y}$$

$$\mathbf{2. \ \ xy=x-y}$$

$$xy = x-y$$ $$\frac{d}{dx} \big[ xy \big] = \frac{d}{dx} \big[ x-y \big]$$ $$1 \cdot y + x \frac{dy}{dx} = 1-\frac{dy}{dx}$$ $$y+x \frac{dy}{dx} = 1 – \frac{dy}{dx}$$ $$x \frac{dy}{dx} + \frac{dy}{dx} = 1-y$$ $$\frac{dy}{dx} \big[ x+1 \big] = 1-y$$ $$\frac{dy}{dx} = \frac{1-y}{x+1}$$

$$\mathbf{3. \ \ x^2-4xy+y^2=4}$$

$$x^2-4xy+y^2=4$$ $$\frac{d}{dx} \big[ x^2-4xy+y^2 \big] = \frac{d}{dx} \big[ 4 \big]$$ $$2x \ – \bigg[ 4x \frac{dy}{dx} + 4y \bigg] + 2y \frac{dy}{dx} = 0$$ $$2x \ – 4x \frac{dy}{dx} – 4y + 2y \frac{dy}{dx} = 0$$ $$-4x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x+4y$$ $$\frac{dy}{dx} \big[ -4x+2y \big] = -2x+4y$$ $$\frac{dy}{dx}=\frac{-2x+4y}{-4x+2y}$$ $$\frac{dy}{dx}=\frac{-x+2y}{-2x+y}$$

$$\mathbf{4. \ \ \sqrt{x+y}=x^4+y^4}$$

$$\sqrt{x+y}=x^4+y^4$$ $$\big( x+y \big)^{\frac{1}{2}}=x^4+y^4$$ $$\frac{d}{dx} \bigg[ \big( x+y \big)^{\frac{1}{2}}\bigg] = \frac{d}{dx}\bigg[x^4+y^4 \bigg]$$ $$\frac{1}{2} \big( x+y \big) ^{-\frac{1}{2}} \bigg( 1+\frac{dy}{dx} \bigg)=4x^3+4y^3\frac{dy}{dx}$$ $$\frac{1}{2} \cdot \frac{1}{\sqrt{x+y}} \cdot \frac{1+\frac{dy}{dx}}{1} = 4x^3+4y^3\frac{dy}{dx}$$ $$\frac{1+\frac{dy}{dx}}{2 \sqrt{x+y}}= 4x^3+4y^3\frac{dy}{dx}$$ $$1+\frac{dy}{dx}= \bigg[ 4x^3+4y^3\frac{dy}{dx} \bigg] \cdot 2 \sqrt{x+y}$$ $$1+\frac{dy}{dx}= 8x^3 \sqrt{x+y} + 8y^3 \frac{dy}{dx} \sqrt{x+y}$$ $$\frac{dy}{dx} \ – \ 8y^3 \frac{dy}{dx} \sqrt{x+y}= 8x^3 \sqrt{x+y} \ – \ 1$$ $$\frac{dy}{dx} \bigg[ 1 \ – \ 8y^3 \sqrt{x+y} \bigg]= 8x^3 \sqrt{x+y} \ – \ 1$$ $$\frac{dy}{dx}= \frac{8x^3 \sqrt{x+y} \ – \ 1}{1 \ – \ 8y^3 \sqrt{x+y}}$$

$$\mathbf{5. \ \ e^{x^2y}=x+y}$$

$$e^{x^2y}=x+y$$ $$\frac{d}{dx} \Big[ e^{x^2y} \Big] = \frac{d}{dx} \big[ x+y \big]$$ $$e^{x^2y} \bigg( 2xy + x^2 \frac{dy}{dx} \bigg) = 1 + \frac{dy}{dx}$$ $$2xye^{x^2y} + x^2e^{x^2y} \frac{dy}{dx} = 1+ \frac{dy}{dx}$$ $$x^2e^{x^2y} \frac{dy}{dx} \ – \ \frac{dy}{dx} = 1 \ – \ 2xye^{x^2y}$$ $$\frac{dy}{dx} \big(x^2e^{x^2y} \ – \ 1 \big) = 1 \ – \ 2xye^{x^2y}$$ $$\frac{dy}{dx} = \frac{1 \ – \ 2xye^{x^2y}}{x^2e^{x^2y} \ – \ 1}$$