Animated Solution for Mathematics - Differentiation: The slope of the tangent to the curve $(y-x^5{)}^2=x(1+x^2{)}^2$ at the point $(1,3)$ is

Welcome, future engineer. Today, we are not just solving a calculus problem; we are peeling back the layers of a geometric mystery. We are presented with the curve defined by the equation $(y-x^5{)}^2=x(1+x^2{)}^2$.

At first glance, this looks like a tangled mess of variables. But in the world of JEE Advanced, complexity is often just a mask for elegance. Our goal is to find the slope of the tangent line at the point $(1,3)$. Remember, the slope of a tangent is the derivative, $\frac{dy}{dx}$.

Many students see an equation like this and immediately panic, trying to isolate $y$. They might try to take the square root of both sides, leading to $y-x^5=\pm \sqrt{x}(1+x^2)$. This is a trap.

It creates two separate branches of the curve and makes the differentiation process unnecessarily tedious. Instead, we use the 'Swiss Army Knife' of calculus: Implicit Differentiation.

We treat $y$ as a function of $x$ and differentiate the entire equation with respect to $x$. This allows us to find the derivative without ever needing to explicitly define $y$ in terms of $x$.

Let's tackle the left-hand side (LHS) first: $\frac{d}{dx}[(y-x^5{)}^2]$. Using the Chain Rule, we bring the power of $2$ down and multiply by the derivative of the inner function.

This gives us:

Now, for the right-hand side (RHS): $\frac{d}{dx}[x(1+x^2{)}^2]$. Here, we have a product of two functions, $x$ and $(1+x^2{)}^2$. We must use the Product Rule: $u^{\prime }v+u{v}^{\prime }$.

Let $u=x$ and $v=(1+x^2{)}^2$. The derivative $u^{\prime }$ is $1$. The derivative $v^{\prime }$ requires the Chain Rule again: $2(1+x^2)\cdot (2x)=4x(1+x^2)$.

Putting it all together, the RHS derivative becomes:

Simplifying this, we get:

Strategic Substitution

Now, we have the full differentiated equation:

Most students would now spend precious minutes trying to isolate $y^{\prime }$. Don't fall for it! We already know the point of tangency is $(1,3)$. Let's substitute $x=1$ and $y=3$ immediately.

On the LHS, substituting $x=1$ and $y=3$ gives us:

On the RHS, substituting $x=1$ gives us:

We are left with the beautifully simple equation:

Dividing by $4$, we get $y^{\prime }-5=3$. Adding $5$ to both sides, we find $y^{\prime }=8$.

The slope of the tangent at $(1,3)$ is exactly $8$. This means that at this specific point, for every unit you move to the right, the curve rises by $8$ units. It is a steep, powerful climb.