Question:

The subtangent at $x = \pi/2$ on the curve $y = x \sin x $ is

Updated On: May 12, 2024
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  • $\pi/2$
  • none of these
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The Correct Option is C

Solution and Explanation

$y=x\,sin\,x$
$\Rightarrow \frac{dy}{dx}=x\,cos\,x+sin\,x\cdot1=x\,cos\,x+sin\,x$
At $x=\frac{\pi}{2}, \frac{dy}{dx}=0+sin \frac{\pi}{2}=1$
and $y=\frac{\pi}{2} sin \frac{\pi}{2} =\frac{\pi}{2}$
Now subtangent $=\left|\frac{y}{y_{1}}\right|_{at\,x=\pi/2}=\frac{\pi/2}{1}=\frac{\pi}{2}$
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives