**Implicit differentiation** is a very powerful technique in differential calculus. It allows us to find derivatives when presented with equations and functions like those in the box. →

One could solve for **y** and find **y'(x)**, but there's an easier way, and it applies to the derivatives of more complicated functions, too.

Implicit differentiation is really just application of the chain rule, where we recognize **y** as a function of **x**, and further differentiate any term containing **y** using the chain rule. For example,

$$\frac{d}{dx} \, y^n = n y^{n - 1} \frac{dy}{dx}$$

These equations all have graphs for which we might want to know the slope.

$$ \begin{align} x^2 + y^2 &= 4 \\[5pt] x^3 + y^3 &= 8 \\[5pt] (x - y)^2 &= x + y + 7 \\[5pt] \sqrt{x^2 + y^2} &= x \end{align}$$

The best way to learn implicit differentiation is by example, so here we go ...

If $x^2 + y^2 = 4$, find $\frac{dy}{dx}$

It's possible to solve for **y** in this equation, of course, and then find **dy/dx**, but implicit differentiation makes finding the derivative much easier.

We start by taking the derivative with respect to **x** (we could as easily take it with respect to **y**) of each term on both sides. We apply the sum rule (the derivative of a sum is the sum of derivatives) on the left and recall that the derivative of a constant is zero.

$$\frac{d}{dx} (x^2 + y^2) = \frac{d}{dx} 4$$

Here is an intermediate step:

$$\frac{d}{dx} x^2 + \frac{d}{dx} y^2 = 0$$

To differentiate **y ^{2}**, we treat

$$2x + 2y \frac{dy}{dx} = 0$$

Moving the **2x** to the right by subtraction and dividing by **2y**, gives us the solution:

$$\frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y}$$

Does the derivative make sense? The graph of this function is a circle centered at the origin, so we expect a positive slope in the 2^{nd} and 4^{th} quadrants, and a negative slope in the 1^{st} and 3^{rd}, just what we get from this derivative. Here's a picture of that:

If $(x - y)^2 = x + y + 7,$ find $\frac{dy}{dx}$

To find ^{dy}**/ _{dx}**, we take the derivative of both sides with respect to

$$\frac{d}{dx} (x - y)^2 = \frac{d}{dx} (x + y + 7)$$

We need the chain rule on both the left and the right is just the derivative of a sum that will include **dy/dx**, remembering that **y** is implicitly a function of **x**:

$$2(x - y)\left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx} + 0$$

Now we'll begin cleaning up and rearranging to solve for ^{dy}**/ _{dx}**:

$$(2x - 2y) \left( 1 - \frac{dy}{dx} \right) = 1 + \frac{dy}{dx}$$

Multiplying the two binomials on the left gives

$$2x - 2x\frac{dy}{dx} - 2y + 2y \frac{dy}{dx} = 1 + \frac{dy}{dx} $$

Now we'll collect all terms containing ^{dy}**/ _{dx}** on the left, and factor out the

$$(2y - 2x - 1)\frac{dy}{dx} = 1 - 2x + 2y$$

Finally, it's just a matter of dividing by **2y - 2x - 1** on both sides to find the derivative:

$$\frac{dy}{dx} = \frac{1 - 2x + 2y}{-1 - 2x + 2y}$$

Once again, we could isolate **y** as a function of **x** and just take a straightforward derivative, but solving for **y** is not quite so easy. It would involve completing the square on **y**, then we'd be left with a tricky derivative. Implicit differentiation is much easier.

If $y = sin^{-1}(x)$, find $\frac{dy}{dx}$

Now we'll see how implicit differentiation can be a powerful tool for solving problems that we might not otherwise be able to, like finding the derivative of **y = sin ^{-1}(x)**.

The first step is to take the sine of both sides, which 'cancels' the inverse sine on the right:

$$sin(y) = sin(sin^{-1}(x)) = x$$

Then take the derivative with respect to x of each side:

$$\frac{d}{dx} sin(y) = \frac{d}{dx} x$$

Now y is implicitly a function of **x**, so we have in instance of ^{dy}**/ _{dx}**:

$$cos(y) \frac{dy}{dx} = 1$$

Now it's just a matter of solving for ^{dy}**/ _{dx}**, as we've done before:

$$\frac{dy}{dx} = \frac{1}{cos(y)}$$

This solution isn't really what we want, however, because it's a function of **y**. To make it a function of **x**, we construct the triangle for which **sin(y) = x**.

Now we can replace **cos(y)** with the bottom side of the triangle in the figure:

$$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$

We can derive a similar formula for the derivative of the inverse cosine function.

Determine the equation of the line tangent to the graph of $y^4 + xy = 2x^3 - 7x - 7$ at the point *(1, 1)*

First we will find ^{dy}**/ _{dx}** by implicit differentiation, then plug

The equation is

$$y^4 + xy = 2x^3 - 7x + 7$$

We need the slope, ^{dy}**/ _{dx}**, so we first differentiate the sums on each side of the equal sign, making sure to treat

$$\frac{d}{dx} y^4 + \frac{d}{dx} xy = \frac{d}{dx}(2x^3 - 7x + 7)$$

We have to use the product rule to do the second derivative on the left, and the result is:

$$4y \frac{dy}{dx} + y + x\frac{dy}{dx} = 6x^2 - 7$$

Now if we gather terms in ^{dy}**/ _{dx}** on the left, we get

$$(4y^3 + x) \frac{dy}{dx} = 6x^2 - 7x - y$$

Dividing by **4y ^{3} + x** on both sides gives us our derivative, which we can evaluate at (1, 1) to get the slope there:

$$\frac{dy}{dx} = \frac{6x^2 - 7 - y}{4y^3 + x} \bigg|_{(1, 1)} = \frac{-2}{5}$$

From there it's simple to use the slope and the point (1, 1) to find the equation of the tangent line. It's

$$2x + 5y = 7$$

Here's a graph of the equation and the tangent line. You can see that implicit differentiation allows us to do things with crazy, non-function equations that we might not have thought possible before.

Find the locations of all vertical and horizontal tangents of the curve described by the curve $x + xy + 2x^3 = 2$

Our goal will be to find the derivative of the equation with respect to **x**, then find the zeros of its numerator (horizontal tangents) and the zeros of its denominator (vertical tangents). The function is

$$x + xy + 2x^3 = 2$$

If we take the implicit derivative, making sure to use the product rule on the second term (**xy**), we get:

$$1 + x \frac{dy}{dx} + y + 6x^2 = 0$$

Solving for dy/dx gives us the derivative, a function of x and y.

$$\frac{dy}{dx} = \frac{-6x^2 - y - 1}{x}$$

Sometimes that's not too convenient. In this case, we can, however, solve the original equation for **y**:

$$y = \frac{-2x^3 - x + 2}{x}$$

Dividing each term of the numerator by **x** gives us:

$y = -2x^3 - 1 + \frac{2}{x}$

We can take the derivative of **y(x)** directly now to get

$$\frac{dy}{dx} = -4x - \frac{2}{x^2}$$

If we insert equation

we get:

$$\frac{dy}{dx} = \frac{1}{x}(-6x^2 + 2x^2 + 1 - \frac{2}{x} - 1)$$

which simplifies to this:

$$\frac{dy}{dx} = \frac{1}{x \left( -4x^2 - \frac{2}{x} \right)}$$

and finally dividing by **x** gives us the same derivative:

$$\frac{dy}{dx} = -4x - \frac{2}{x^2}$$

Now finding the horizontal tangent(s) means setting that derivative equal to zero:

$$-4x - \frac{2}{x^2} = 0$$

Moving one term to the other side gives us:

$$-4x = \frac{2}{x^2}$$

Multiplying both sides by **x ^{2}** and simplifying the constant gives us a cubic equaiton that can be solved

$$x^3 = -\frac{1}{2}$$

The result is the location of a single horizontal tangent at

$$x = -0.794 \\ y = 2.779$$

Vertical tangent(s) occur when the denominator of the derivative is zero. Taking the implicit derivative, we see that it goes to zero when **x** goes to zero:

Here is a graph of the function showing the single horizontal tangent and the vertical tangent, at which the function has no value.

Find the derivative of each with respect to x.

1. | $3 y^3 + x^2 = 5$ | |

2. | $y^4 - 2y = 4x^3 + x$ | |

3. | $x^2y + 2x^3 y = x + y$ | |

4. | $xy^2 + x^2y^5 - x^3 = 3$ | |

5. | $x^3 y^5 = 1$ | |

6. | $x^4 + y^4 = 1$ | |

7. | $\frac{y}{x} + \frac{x}{y} = 2y$ |

8. | $\sqrt{x + y} = \frac{1}{x} + \frac{1}{y}$ | |

9. | $y^{-2/3} + x^{3/2} = 1$ | |

10. | $x^{3/2} + y^{1/3} = -5y$ | |

11. | $y + \frac{1}{y} = x^3 + x$ | |

12. | $sin(xy) = x$ | |

13. | $sin(x + y) = x + cos(y)$ | |

14. | $tan(x^2 y) = (x + y)^2$ |

Implicit differentiation can be the best route to what otherwise could be a tricky differentiation. In example 3 above we found the derivative of the inverse sine function. We can use that as a general method for finding the derivative of any inverse function.

Recall that if **f(x)** is a function and **f ^{-1}(x)** is its inverse, that

In other words, a function and its inverse "undo" one another.

To find the derivative of an inverse function, begin with

$$f(f^{-1}(x)) = x$$

Now differentiate both sides with respect to x, but be aware that the left side is a composition of functions, and will require the chain rule:

$$\frac{d}{dx} \, f(f^{-1}(x)) = \frac{d}{dx} \, x$$

The chain rule gives us:

$$f'(f^{-1}(x)) \cdot f'^{-1}(x) = 1$$

where **f ^{'-1}(x)** is the derivative of the inverse. Solving for the derivative of the inverse, we get:

$$\frac{d}{dx} f^{-1} (x) = \frac{1}{f'(f^{-1}(x))}$$

So to find the derivative of the inverse of the function at a point **x**, we substitute the inverse function for **x** in **f'(x)**, then take the reciprocal.

**g(x)** instead of **f ^{-1}(x)** in order to avoid confusion of the primes and -1’s in the exponent. Make that substitution if you need to. I prefer always to refer to the inverse function as

The derivative of the inverse of a function is equal to the reciprocal of the compound function **f'(f ^{-1}(x))**. We need to know the inverse function and the derivative of

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