Animated Solution for Mathematics - Differentiation: The function $f(x)=\frac{\ln (\pi +x)}{\ln (e+x)}$ is

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a function that, at first glance, looks like a simple ratio of logarithms:

It is easy to feel intimidated by such expressions, but remember: calculus is not about memorizing formulas; it is about understanding the heartbeat of change. We want to know if this function climbs or slides as $x$ moves from $0$ to $\infty$.

Our Compass

To determine if a function is increasing or decreasing, we look at its derivative, $f^{\prime }(x)$. Think of the derivative as the slope of the tangent line at any point $x$.

If $f^{\prime }(x)>0$, the function is climbing; if $f^{\prime }(x)<0$, it is sliding down. Our mission is to find the sign of this derivative.

Since our function is a quotient, we invoke the trusty Quotient Rule:

where $u=\ln (\pi +x)$ and $v=\ln (e+x)$.

Let's perform the differentiation with care. The derivative of $\ln (\pi +x)$ is $\frac{1}{\pi +x}$, and the derivative of $\ln (e+x)$ is $\frac{1}{e+x}$.

Plugging these into our rule, we get:

Take a deep breath. I know this looks like a complex algebraic mess, but look at the denominator: $[\ln (e+x){]}^2$.

As we discussed, for $x>0$, $e+x>1$, so the log is positive, and its square is definitely positive. The denominator is just a spectator; the real drama happens in the numerator.

Now, we must compare the two terms in the numerator:

We know that $\pi \approx 3.14$ and $e\approx 2.71$. Clearly, $\pi >e$.

This implies $\pi +x>e+x>0$. When we take the reciprocals, the inequality flips:

Furthermore, because the natural logarithm is an increasing function, $\ln (\pi +x)>\ln (e+x)$.

Look at $T_1$ and $T_2$ again. $T_1$ is a smaller fraction multiplied by a smaller log value. $T_2$ is a larger fraction multiplied by a larger log value.

When you subtract a larger product from a smaller product, the result is inevitably negative. Thus, the numerator is strictly less than zero.

Since the numerator is negative and the denominator is positive, the derivative $f^{\prime }(x)$ is negative for all $x>0$.

Geometrically, this means the slope of our function is always pointing downwards. The function $f(x)$ is strictly decreasing on the interval $(0,\infty )$.

We have successfully navigated the complexity and found the truth hidden within the algebra. Keep practicing, keep questioning, and keep falling in love with the elegance of mathematics!