Question

The tangent of the curve $y=\sin \left(\sqrt{\cos x}\right)$ at $x=\frac{\mathrm{\pi }}{2}$ has gradient

• A. $0$
• B. $\infty$
• C. $-\infty$
• D. $-\frac{1}{2}$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Step 1: Differentiate the given curve

Given, $y=\sin \left(\sqrt{\cos x}\right)$

The slope of the tangent of a curve is described by its derivative.

The derivative of the given function is,

$\begin{array}{ccc}& \frac{dy}{dx}=\frac{\mathrm{d}}{dx}\sin \left(\sqrt{\cos x}\right)& \\ \Rightarrow & \frac{dy}{dx}=\cos \left(\sqrt{\cos x}\right)·\frac{1}{2\sqrt{\cos x}}\left(-\sin x\right)& ...\left[\because \frac{d}{dx}f\left(g\left(x\right)\right)=f\text{'}\left(g\left(x\right)\right)·g\text{'}\left(x\right)\right]\\ \Rightarrow & \frac{dy}{dx}=-\frac{\cos \left(\sqrt{\cos x}\right)\sin x}{2\sqrt{\cos x}}& \end{array}$

Step 2: Evaluate the derivative at $x=\frac{\mathrm{\pi }}{2}$

To evaluate the gradient of a function at a given point, we have to substitute the value of the $x$-coordinate at the point into the derivative of the function.

But directly substituting $x=\frac{\mathrm{\pi }}{2}$ yields indeterminate form. Thus, we have to take the limit of the derivative at $x=\frac{\mathrm{\pi }}{2}$.

Thus,

$\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left(\frac{dy}{dx}\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[-\frac{\cos \left(\sqrt{\cos x}\right)\sin x}{2\sqrt{\cos x}}\right]$

Applying L'Hospital's rule,

$\begin{array}{rcl}\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left(\frac{dy}{dx}\right)& =& \underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[-\frac{\cos \left(\sqrt{\cos x}\right)\sin x}{2\sqrt{\cos x}}\right]\\ & =& \underset{x\to \frac{\mathrm{\pi }}{2}}{l\mathrm{im}}\left[\frac{\sin \left(\sqrt{\cos x}\right)·{\displaystyle \frac{1}{2\sqrt{\cos x}}}·{\sin }^{2}x-\cos x·\cos \left(\sqrt{\cos x}\right)}{2{\displaystyle \frac{1}{2\sqrt{\cos x}}}·\left(-\sin x\right)}\right]\\ & =& \underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{\sin \left(\sqrt{\cos x}\right)·{\sin }^{2}x-2\sqrt{\cos x}·\cos x·\cos \left(\sqrt{\cos x}\right)}{\left(-\sin x\right)}\right]\\ & =& \left[\frac{\sin \left(\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}\right)·{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)-2\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}·\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)·\cos \left(\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}\right)}{\left(-\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)\right)}\right]\\ & =& \left[\frac{\sin \left(0\right)·1-2·0·0·\cos \left(0\right)}{-1}\right]\\ & =& 0\end{array}$

Hence, option A is correct.