# Unlocking the Secrets of Inverse Functions: A Closer Look

Finding the derivative of an inverse function involves flipping the derivative of the original function, mirroring changes inversely. For trigonometric inverses, like arcsin or arctan, it's about understanding how angles change in circular motion. This process requires a keen grasp of how functions and their inverses are interlinked, reflecting each other's changes in a reciprocal manner.

## How to find the derivative of inverse function?

The derivative of an inverse function is computed by taking the reciprocal of the derivative of the original function, considering the interdependent relationship between a function and its inverse. For trigonometric inverses, such as arcsin or arctan, the process involves understanding the derivatives of circular functions and their inverse relationship, focusing on the rate of change of angles with respect to their trigonometric ratios.

You can find the inverse of a function first and then compute its derivative. This approach involves two main steps: first, determine the inverse function by reversing the roles of the dependent and independent variables in the original function, and then solve for the new dependent variable. Once you have the inverse function explicitly, you can apply standard differentiation techniques to find its derivative.

However, this method can be challenging if finding the inverse function explicitly is difficult or if the inverse function is complex, making differentiation tricky. In such cases, using the formula that relates the derivatives of a function and its inverse can be more efficient and straightforward:

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

Here are the derivatives of the inverse trigonometric functions, with a variable $$(x)$$ or a function $$(f(x))$$ inside of them:

### Sin:

$$\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 – x^2}}$$

$$\frac{d}{dx} \arcsin(f(x)) = \frac{f'(x)}{\sqrt{1 – f(x)^2}}$$

### Cos:

$$\frac{d}{dx} \arccos(x) = \frac{-1}{\sqrt{1 – x^2}}$$

$$\frac{d}{dx} \arccos(f(x)) = \frac{-f'(x)}{\sqrt{1 – f(x)^2}}$$

### Tan:

$$\frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2}$$

$$\frac{d}{dx} \arctan(f(x)) = \frac{f'(x)}{1 + f(x)^2}$$

### Sec:

$$\frac{d}{dx} \text{arcsec}(x) = \frac{1}{|x| \sqrt{x^2 – 1}}$$

$$\frac{d}{dx} \text{arcsec}(f(x)) = \frac{f'(x)}{|f(x)| \sqrt{f(x)^2 – 1}}$$

Example 1:

Consider the function $$f(x) = 2x$$ and its derivative $$f'(x) = 2$$

The derivative of $$arcsin f(x)$$ is given by:

$$\frac{d}{dx} \arcsin(f(x)) = \frac{f'(x)}{\sqrt{1 – f(x)^2}}$$

By substituting the values, we have: $$\frac{2}{\sqrt{1 – (2x)^2}} = \frac{2}{\sqrt{1 – 4x^2}}$$

Example 2:

Given the function $$f(x) = \frac{1}{3}x^3$$, find the derivative of its inverse at the point where $$x = \frac{1}{3}$$

Solution: its derivative is $$f'(x) = x^2$$.

To find where $$x = \frac{1}{3}$$ in the original function: $$\frac{1}{3}x^3 = \frac{1}{3}$$

This gives $$x = 1$$, as the point in the original function corresponding to$$x = \frac{1}{3}$$

in the inverse function.

The derivative of the inverse function is given by:

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

Substituting $$x=1$$ into the derivative of the original function, we get $$f'(1) = 1^2 = 1$$

Thus, the derivative of the inverse function at $$x = \frac{1}{3}$$ is:

$$\frac{1}{f'(f^{-1}(\frac{1}{3}))} = \frac{1}{1} = 1$$

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