Animated Solution for Mathematics - Differentiation: The value of ${\log }_{e}2\frac{d}{dx}({\log }_{\cos x}\text{cosec}x)$ at $x=\frac{\pi }{4}$ is

Welcome, future engineer. Today, we are not just solving a problem; we are dissecting a classic trap. When you look at the expression ${\log }_{\cos x}\text{cosec}x$, your first instinct might be to panic.

It looks like a standard logarithm, but the base is a function of $x$. In the world of JEE Advanced, this is a signal that the standard rules of differentiation—the ones you use for $x^n$ or $e^x$—are not enough.

We need to build a bridge to a more comfortable territory: the natural logarithm, base $e$.

We begin by defining our function $y={\log }_{\cos x}\text{cosec}x$. The problem here is the base, $\cos x$.

To tame this, we invoke the Change of Base Formula: ${\log }_{a}b=\frac{\ln b}{\ln a}$. By applying this, we transform our function into:

Suddenly, the landscape changes. We are no longer dealing with an exotic base; we are dealing with a quotient of two familiar functions. This is the first victory in our journey.

Before we rush into the quotient rule, let us pause. Mathematics rewards those who simplify before they calculate.

We know that $\text{cosec}x=(\sin x{)}^{-1}$. Using the logarithmic property $\ln (a^b)=b\ln a$, we can rewrite the numerator:

Now, our function $y$ looks much cleaner:

This small act of simplification is the difference between a smooth solution and a page full of messy algebra.

Now, we engage the engine: the quotient rule. We want to find $\frac{dy}{dx}$. The rule tells us that for a function $\frac{u}{v}$, the derivative is $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$.

Let $u=\ln (\sin x)$ and $v=\ln (\cos x)$. We keep the negative sign outside to avoid confusion.

Differentiating $u$ gives us $\frac{1}{\sin x}\cdot \cos x=\cot x$. Differentiating $v$ gives us $\frac{1}{\cos x}\cdot (-\sin x)=-\tan x$.

With these pieces in hand, we assemble the derivative:

We are almost at the finish line. The problem asks for the value at $x=\frac{\pi }{4}$. At this specific angle, the trigonometric functions are beautifully symmetric:

Substituting these values, the expression collapses. The numerator becomes $1\cdot \ln (\frac{1}{\sqrt{2}})+1\cdot \ln (\frac{1}{\sqrt{2}})$, which is $2\ln (\frac{1}{\sqrt{2}})$.

The denominator is $(\ln (\frac{1}{\sqrt{2}}){)}^2$. Canceling one term, we are left with:

Since $\ln (\frac{1}{\sqrt{2}})=\ln (2^{-1/2})=-\frac{1}{2}\ln 2$, the expression simplifies to:

The question asks for $\ln 2\cdot \frac{dy}{dx}$. Multiplying our result by $\ln 2$, we get:

The terms cancel out with satisfying precision. You have navigated the variable base, the quotient rule, and the trigonometric evaluation.

This is the essence of JEE Advanced mathematics: not just calculation, but the strategic application of tools to reveal a simple, elegant truth. The final answer is 4.