# What is the derivative of an inverse function?

Asked By: Sarwar Kiernan | Last Updated: 7th April, 2020

Category:
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space and astronomy

The

**Derivative of an Inverse Function**. (f−1)′(a)=pq. f′(f−1(a))=qp. (f−1)′(a)=1f′(f−1(a)).Hereof, how do you find the derivative of an inverse function?

**Finding the Inverse of a Function**

- First, replace f(x) with y .
- Replace every x with a y and replace every y with an x .
- Solve the equation from Step 2 for y .
- Replace y with f−1(x) f − 1 ( x ) .
- Verify your work by checking that (f∘f−1)(x)=x ( f ∘ f − 1 ) ( x ) = x and (f−1∘f)(x)=x ( f − 1 ∘ f ) ( x ) = x are both true.

Similarly, what is the inverse function rule? In order for a **function** to have an **inverse**, it must pass the horizontal line test!! Horizontal line test If the graph of a **function** y = f(x) is such that no horizontal line intersects the graph in more than one point, then f has an **inverse function**.

Similarly, you may ask, how are the derivatives of inverse functions related?

**Derivatives of inverse functions**. **Functions** f and g are inverses if f(g(x))=x=g(f(x)). For every pair of such **functions**, the **derivatives** f' and g' have a special relationship. Learn about this relationship and see how it applies to ??ˣ and ln(x) (which are **inverse functions**!).

What is the derivative of tan 1?

Expression | Derivatives |
---|---|

y = cos^{-}^{1}(x / a) | dy/dx = - 1 / (a^{2} - x^{2})^{1}^{/}^{2} |

y = tan^{-}^{1}(x / a) | dy/dx = a / (a^{2} + x^{2}) |

y = cot^{-}^{1}(x / a) | dy/dx = - a / (a^{2} + x^{2}) |

y = sec^{-}^{1}(x / a) | dy/dx = a / (x (x^{2} - a^{2})^{1}^{/}^{2}) |