Inverse functions are very important in Mathematics as well as in many applied areas of science. The most famous pair of functions inverse to each other are the logarithmic and the exponential functions. Other functions like the tangent and arctangent play also a major role. A mathematical function is an expression that denotes a relationship between inputs and outputs or the independent and dependent variables. The concept of inverse function is quite frequently seen in mathematics. An inverse function "undoes" a particular function. We may say that if the reverse of a function exists, then it is known as its inverse function. Also, the function whose inverse exists, is called an invertible function. Let us suppose that f be a function which has x as input and y as output, then its inverse function (say g) should have y as input and results x; i.e. Let f(x) = y, then g will be the inverse function, if: g(y) = x Here, f is invertible function. Inverse of a function is represented by $f^{-1}$. Thus, in place of g, we may also use $f^{-1}$; i.e. here, $g\ =\ f^{-1}$ The relation between a function f and its inverse $^{-1}$ is: $f(f^{-1}(x))=x=f^{-1}(f(x))$ |

As we discussed that inverse of a function is something (some function) which undoes or reverses the effect of given function. We can say that if inverse of a function f be $f^{-1}$, then we have $y=f(x)⇔x=f^{-1}(y)$

Let us suppose that f be a real-valued invertible function, then the derivative of its inverse function $f^{-1}$ is represented by the following explicit relation:

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

The fact is that the product of the derivatives of function and its inverse is one. We can say that the derivative of function is reciprocal of the derivative of its inverse.

Some important formulas of DIF, which can be utilized for finding derivatives of other functions are given below, have a look.

**a)**$\frac{d}{dx}$ $sin^{-1}x$ = $\frac{1}{\sqrt{1-x^{2}}}$

**b)**$\frac{d}{dx}$ $cos^{-1}x$ = $-\frac{1}{\sqrt{1-x^{2}}}$

**c)**$\frac{d}{dx}$ $tan^{-1}x$ = $\frac{1}{1+x^{2}}$

**d)**$\frac{d}{dx}$ $cot^{-1}x$ = $-\frac{1}{1+x^{2}}$

**e)**$\frac{d}{dx}$ $sec^{-1}x$ = $\frac{1}{x\sqrt{x^{2}-1}}$

**f)**$\frac{d}{dx}$ $csc^{-1}x$ = $-\frac{1}{x\sqrt{x^{2}-1}}$

The solution of differentiation of inverse function have two methods, Let us have a look.

**Method 1**

**Step 1:**Let the given function be f(x). Assume that y = f(x).

For Example: In order to find derivative of inverse of f(x) = x + 1, write y = x + 1.

**Step 2:**Determine the inverse function; i.e. find the value of x in terms of y.

y = x + 1

x = y - 1

**Step 3:**Now, replace x by y and y by x.

i.e. y = x - 1

which gives inverse function of given function.

**Step 4:**The last and final step is to find its derivative

$\frac{dy}{dx}$ = $1\ - 0\ =\ 1$

**Method 2**

**Step 1:**Let y = f(x)

Find the inverse function; i.e. calculate the value of x in terms of y.

$f(y)\ =\ x$

**Step 2:**Differentiate both sides using chain rule:

$\frac{df}{dx}$ $\cdot$ $\frac{dy}{dx}$ = $1$

**Step 3:**Find the value of $\frac{dy}{dx}$

i.e.

$\frac{dy}{dx}$ = $\frac{1}{\frac{df}{dy}}$

**Step 4:**Use the relation $f(f^{-1}(x))\ =\ x\ =\ f^{-1}(f(x))$ when required.

Some of the solved examples of derivative of inverse of the function are given below:

**Example 1:**

Find the derivative of the inverse of 4x + 3.

**Solution:**

Let y = 4x + 3

$x$ = $\frac{1}{4}$ $(y−3)$

Replace x by y and y by x

$y$ = $\frac{1}{4}$ $(x−3)$

Differentiate with respect to x,

$\frac{dy}{dx}$ = $\frac{1}{4}$ $(1−0)$

$\frac{dy}{dx}$ = $\frac{1}{4}$

**Example 2:**

Find the derivative of $cos^{-1}x$.

**Solution:**

Let $y = cos^{-1}x$

cos y = x

Differentiate with respect to x

$-sin\ y\ .$ $\frac{dy}{dx}$ = $1$

$\frac{dy}{dx}$ = $-\frac{1}{sin y}$

$\frac{dy}{dx}$ = $-\frac{1}{sin(cos^{-1x})}$

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

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