Animated Solution for Mathematics - Differentiation: If $y^2+{\log }_e({\cos }^{2}x)=y,x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, then :

Imagine you are standing on the edge of a mathematical cliff, looking at an equation that seems to hide its secrets: $y^2+{\log }_e({\cos }^{2}x)=y$.

In the world of JEE Advanced, the most complex-looking problems often have the most elegant solutions if you know how to peel back the layers. Let's embark on this journey together.

The Logarithmic Simplification

The first rule of battle is to simplify your terrain. We see a logarithmic term, ${\log }_e({\cos }^{2}x)$.

Before we even think about touching a derivative, let's use the power rule of logarithms: ${\log }_e(a^b)=b{\log }_{e}a$. This transforms our equation into:

Suddenly, the expression feels much lighter. We are now ready to engage.

The Initial State at $x=0$

The question asks us to evaluate derivatives at $x=0$. This is a massive hint!

Whenever you see $x=0$ in a trigonometric context, your mind should immediately jump to the values $\cos (0)=1$ and $\tan (0)=0$. Let's substitute $x=0$ into our simplified equation:

Since ${\log }_e(1)=0$, the entire logarithmic term vanishes, leaving us with $y(0{)}^2-y(0)=0$. Factoring this gives us $y(0)(y(0)-1)=0$, meaning $y(0)$ can be either $0$ or $1$. We have two potential paths, and we must respect both.

The First Derivative

Now, we differentiate with respect to $x$. Using implicit differentiation, the derivative of $y^2$ is $2y{y}^{\prime }$, and the derivative of $2{\log }_e(\cos x)$ is:

So, our equation becomes $2y{y}^{\prime }-2\tan x=y^{\prime }$. At $x=0$, since $\tan (0)=0$, this simplifies beautifully to $2y(0)y^{\prime }(0)=y^{\prime }(0)$.

Rearranging, we get $y^{\prime }(0)(2y(0)-1)=0$. Since $y(0)$ is either $0$ or $1$, the term $(2y(0)-1)$ is never zero. Therefore, $y^{\prime }(0)$ must be $0$. The first hurdle is cleared!

The Second Derivative

This is where most students stumble, but you won't. We differentiate $2y{y}^{\prime }-2\tan x=y^{\prime }$ again.

Applying the product rule to $2y{y}^{\prime }$, we get $2(y^{\prime }{)}^2+2y{y}^{\prime \prime }$. The derivative of $-2\tan x$ is $-2{\sec }^{2}x$. So, we have:

Now, substitute $x=0$ and $y^{\prime }(0)=0$. The equation collapses into $0+2y(0)y^{\prime \prime }(0)-2(1{)}^2=y^{\prime \prime }(0)$.

The Final Synthesis

We have two cases for $y(0)$. If $y(0)=0$, then $-2=y^{\prime \prime }(0)$, so $|y^{\prime \prime }(0)|=2$.

If $y(0)=1$, then $2y^{\prime \prime }(0)-2=y^{\prime \prime }(0)$, which means $y^{\prime \prime }(0)=2$, so $|y^{\prime \prime }(0)|=2$.

In both scenarios, the magnitude of the second derivative is $2$. We have conquered the problem! It wasn't about brute force; it was about systematic, step-by-step simplification.