Animated Solution for Mathematics - Differentiation: The smallest positive root of the equation, $\tan x-x=0$ lies in

To find the smallest positive root of the equation $\tan x-x=0$, we recognize that this is a transcendental equation. Standard algebraic manipulation is insufficient here, so we shift our perspective to the Cartesian plane.

We are seeking the intersection of two distinct curves:

Consider the interval $(0,\frac{\pi }{2})$. We define the function $f(x)=\tan x-x$ and examine its rate of change:

Using the fundamental trigonometric identity ${\sec }^{2}x-1={\tan }^{2}x$, we find that $f^{\prime }(x)={\tan }^{2}x$. Since the square of any real number is non-negative, $f^{\prime }(x)\ge 0$, implying the function is strictly increasing.

Given that $f(0)=0$, the function immediately climbs above the $x$-axis for all $x>0$ in this interval. Consequently, the curve $y=\tan x$ remains strictly above the line $y=x$, and no intersection occurs.

Next, we examine the interval $(\frac{\pi }{2},\pi )$. In this region, the tangent function yields negative values.

Conversely, the line $y=x$ remains positive throughout this interval. Because a negative value can never equal a positive value, there is no intersection in this domain.

Finally, we arrive at the interval $(\pi ,\frac{3\pi }{2})$. We apply the Intermediate Value Theorem to determine if a root exists.

At the lower boundary $x=\pi$: This value is clearly negative.

As $x$ approaches $\frac{3\pi }{2}$ from the left, the tangent function approaches positive infinity. Therefore, $f(x)$ becomes positive.

Because the function is continuous on this interval and transitions from a negative value to a positive value, it must cross the $x$-axis. We conclude that the smallest positive root lies within the interval $(\pi ,\frac{3\pi }{2})$.