Animated Solution for Mathematics - Differentiation: $\frac{d^{2}x}{d{y}^2}$ equals

Welcome, future IITians! Today, we are going to tackle a classic calculus problem that often trips students up in the JEE. We want to find the second derivative of $x$ with respect to $y$, written as $\frac{d^{2}x}{d{y}^2}$.

A very common mistake is to think that this is simply the reciprocal of $\frac{d^{2}y}{d{x}^2}$. But calculus doesn't work that way!

To understand why, imagine you have plotted a function $y=f(x)$ in blue, and its inverse function $x=f^{-1}(y)$ in red. They are perfect reflections of each other across the line $y=x$. When you change the perspective from $x$ to $y$, you are not just flipping the curve; you are fundamentally changing how the slope evolves.

Our Foundation

Let's start with the first derivative. At any point $P$ on our blue curve, the slope of the tangent is $\frac{dy}{dx}$. At the corresponding point $Q$ on the red curve, the slope of the tangent with respect to the $x$-axis is $\frac{dx}{dy}$.

By the Inverse Function Theorem, these two slopes are reciprocals of each other. So, we can write:

To make our lives easier for the next steps of differentiation, let's write this as:

This is our starting equation, and it is the bedrock upon which we will build our solution.

Why Reciprocals Fail

Now, we need to find the second derivative, which is $\frac{d^{2}x}{d{y}^2}$. By definition, this is simply the derivative of $\frac{dx}{dy}$ with respect to $y$.

Notice the variable in the denominator—it is $y$, not $x$! If you try to just flip the second derivative, you are ignoring the fact that the rate of change of the slope is dependent on the variable you are moving along.

We must differentiate the expression ${\left(\frac{dy}{dx}\right)}^{-1}$ with respect to $y$. This is where we must be extremely careful because the inner function is in terms of $x$, but we are differentiating with respect to $y$.

The Bridge

Since we cannot directly differentiate a function of $x$ with respect to $y$, we must use our trusty tool: the Chain Rule! The Chain Rule allows us to convert the derivative operator $\frac{d}{dy}$ into $\frac{d}{dx}\cdot \frac{dx}{dy}$.

Let's apply this to our expression. Now, our equation becomes:

This is a crucial step where many students make a silly mistake and forget that extra factor of $\frac{dx}{dy}$!

Now, let's focus entirely on differentiating the term inside the brackets with respect to $x$. We have ${\left(\frac{dy}{dx}\right)}^{-1}$. Using the power rule, the exponent $-1$ comes to the front, and the power decreases by one, becoming $-2$.

But wait, we must also differentiate the inner function, which is $\frac{dy}{dx}$, with respect to $x$! The derivative of $\frac{dy}{dx}$ with respect to $x$ is simply the second derivative, $\frac{d^{2}y}{d{x}^2}$.

Putting it all together, we get:

Now, let's substitute this back into our main chain rule equation. We replace the first part with our newly calculated expression, and we still have that extra factor of $\frac{dx}{dy}$ at the end.

But remember, we want our final answer entirely in terms of derivatives of $y$ with respect to $x$. So, let's replace $\frac{dx}{dy}$ with its equivalent, ${\left(\frac{dy}{dx}\right)}^{-1}$.

Now, look at the equation: we have a product of terms with the same base, $\frac{dy}{dx}$. Finally, we just need to simplify the exponents. Since we are multiplying ${\left(\frac{dy}{dx}\right)}^{-2}$ with ${\left(\frac{dy}{dx}\right)}^{-1}$, we simply add their exponents. $-2+(-1)$ gives us $-3$!

So, our final expression is:

This perfectly matches Option 3! This is a beautiful and highly important result that you should memorize for the JEE, as it saves a lot of time during the exam. Keep practicing, and remember: in calculus, always respect the variable of differentiation!