Subject: Calculus

Inverse Functions And Differentiation

In algebra, we are actually taught about functions and its corresponding inverses. In fact, I would really like to assume here that all of us know about Inverse Functions and proceed directly to differentiation of inverse functions. But for the sake of those who didn't knew about it and to those who knew but they forgot (which constitutes for the majority, I guess), then I would give just a brief introduction regarding inverse functions.

Inverse Functions

Basically, an inverse of a function is something that "undoes" the effect of that function. For example, if we let a device that converts dollars to euro as a function, then its inverse would be a device that converts back euro to dollars. Geometrically, the graph of an inverse function is just a "reflection" of that function (see figure at the right). So as you see in the figure, the plot of the function is the color red and its inverse is colored in blue. Now, if we fold that figure along the black dotted line, the function and its inverse coincides, hence the term reflection.
So, how do we solve the inverse of any given function? Well, since a function is expressed in terms of x (y=f(x)), then its inverse can be solved by expressing the variable x in terms of y. An illustration of this would be,
Find the inverse of the function y=f(x)=x^2.


Find the inverse of the function y=e^{x}.




Differentiation of Inverse Functions

Solving for the inverse of a function was just like playing with algebra so easily. Now, we will do some differentiation on inverse functions. We know that the Leibniz notation for the derivative of a function y=f(x) is \frac{dy}{dx}. Now, since the inverse can be written as x=f(y), then the Leibniz notation for its derivative is automatically \frac{dx}{dy}. Multiplying these two derivatives, we had,

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


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

which only tells us that the derivatives of a function and its inverse are reciprocal. Furthermore, if we write the inverse of f(x) as g(y), then standard derivative notation would agree that the derivative of an inverse, which is \frac{dx}{dy}, can also be written as g'(y). Thus,


and since \frac{dy}{dx}=f'(x), then

\frac{dy}{dx}\cdot \frac{dx}{dy}=f'(x)\cdot g'(y)=1

rearranging, we get


and we know that x=f(y)=g(y), thus


which is just the formula for the derivative of inverse functions. Now lets test our solved inverse functions above if they really are correct ones by testing their reciprocity.
A. The function is f(x)=y=x^2 and its inverse is given by g(y)=\sqrt{y}. Their derivatives are


Then its product is given by

f'(x)\cdot g'(y)=2x\cdot \frac{1}{2}\sqrt{y}

but we know that \sqrt{y}=x, then

f'(x)\cdot g'(y)=2x\cdot \frac{1}{2x}=1

Indeed! They are reciprocal.

Standard Notation for Inverse Functions

If f(x) is a function, then its inverse is standardly denoted as


However, one must not confuse the -1 in the superscript of f as exponents.:-)

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