Question

Let $f(x) = cos^{-1}(2x\sqrt{1-x^2})$ then $f'(\frac{\sqrt{3}}{2})$ equals

• A. -4
• B. 4
• C. $\frac{-1}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

4

Answer 2

To find $f'(\frac{\sqrt{3}}{2})$ for $f(x) = \cos^{-1}(2x\sqrt{1-x^2})$, we can use two methods:

Method 1: Using Trigonometric Substitution

Let $x = \sin\theta$. Since $x = \frac{\sqrt{3}}{2}$ is positive, we can choose $\theta \in [0, \frac{\pi}{2}]$. Then $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = \cos\theta$ (since $\cos\theta \ge 0$ for $\theta \in [0, \frac{\pi}{2}]$).

Substitute $x = \sin\theta$ into $f(x)$: $f(x) = \cos^{-1}(2\sin\theta\cos\theta)$ $f(x) = \cos^{-1}(\sin(2\theta))$

Now, use the identity $\sin A = \cos(\frac{\pi}{2} - A)$: $f(x) = \cos^{-1}(\cos(\frac{\pi}{2} - 2\theta))$

For $x = \frac{\sqrt{3}}{2}$, we have $\sin\theta = \frac{\sqrt{3}}{2}$, which implies $\theta = \frac{\pi}{3}$. Then $2\theta = \frac{2\pi}{3}$. So, $\frac{\pi}{2} - 2\theta = \frac{\pi}{2} - \frac{2\pi}{3} = \frac{3\pi - 4\pi}{6} = -\frac{\pi}{6}$.

Now, we have $f(x) = \cos^{-1}(\cos(-\frac{\pi}{6}))$. The principal value branch of $\cos^{-1}(y)$ is $[0, \pi]$. Since $-\frac{\pi}{6}$ is not in $[0, \pi]$, we use the property $\cos^{-1}(\cos A) = -A$ if $A \in [-\pi, 0]$. Here $A = -\frac{\pi}{6}$, which is in $[-\pi, 0]$. So, $f(x) = -(-\frac{\pi}{6}) = \frac{\pi}{6}$. This is the value of the function at $x=\frac{\sqrt{3}}{2}$, not the derivative.

To find the derivative, we need the functional form of $f(x)$ for the relevant range of $x$. For $x = \frac{\sqrt{3}}{2}$, $\theta = \frac{\pi}{3}$. The angle $\frac{\pi}{2} - 2\theta = -\frac{\pi}{6}$ falls in the range $[-\frac{\pi}{2}, 0)$. For $A \in [-\frac{\pi}{2}, 0)$, $\cos^{-1}(\cos A) = -A$. So, $f(x) = -(\frac{\pi}{2} - 2\theta) = 2\theta - \frac{\pi}{2}$. Substitute back $\theta = \sin^{-1}x$: $f(x) = 2\sin^{-1}x - \frac{\pi}{2}$.

Now, differentiate $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}(2\sin^{-1}x - \frac{\pi}{2})$ $f'(x) = 2 \cdot \frac{1}{\sqrt{1-x^2}} - 0$ $f'(x) = \frac{2}{\sqrt{1-x^2}}$

Finally, substitute $x = \frac{\sqrt{3}}{2}$: $f'(\frac{\sqrt{3}}{2}) = \frac{2}{\sqrt{1-(\frac{\sqrt{3}}{2})^2}} = \frac{2}{\sqrt{1-\frac{3}{4}}} = \frac{2}{\sqrt{\frac{1}{4}}} = \frac{2}{\frac{1}{2}} = 4$.

Method 2: Direct Differentiation using Chain Rule

Let $f(x) = \cos^{-1}(2x\sqrt{1-x^2})$. Using the chain rule, $\frac{d}{dx}\cos^{-1}(u) = \frac{-1}{\sqrt{1-u^2}}\frac{du}{dx}$. Here, $u = 2x\sqrt{1-x^2}$.

First, find $\frac{du}{dx}$ using the product rule: $u = 2x(1-x^2)^{1/2}$ $\frac{du}{dx} = 2 \left[ (1)\cdot(1-x^2)^{1/2} + x \cdot \frac{1}{2}(1-x^2)^{-1/2}(-2x) \right]$ $\frac{du}{dx} = 2 \left[ \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} \right]$ $\frac{du}{dx} = 2 \left[ \frac{(1-x^2) - x^2}{\sqrt{1-x^2}} \right] = \frac{2(1-2x^2)}{\sqrt{1-x^2}}$.

Now, evaluate $\frac{du}{dx}$ at $x = \frac{\sqrt{3}}{2}$: $\frac{du}{dx} \Big|_{x=\frac{\sqrt{3}}{2}} = \frac{2(1-2(\frac{\sqrt{3}}{2})^2)}{\sqrt{1-(\frac{\sqrt{3}}{2})^2}} = \frac{2(1-2\cdot\frac{3}{4})}{\sqrt{1-\frac{3}{4}}} = \frac{2(1-\frac{3}{2})}{\sqrt{\frac{1}{4}}} = \frac{2(-\frac{1}{2})}{\frac{1}{2}} = \frac{-1}{\frac{1}{2}} = -2$.

Next, evaluate $u$ at $x = \frac{\sqrt{3}}{2}$: $u = 2(\frac{\sqrt{3}}{2})\sqrt{1-(\frac{\sqrt{3}}{2})^2} = \sqrt{3}\sqrt{1-\frac{3}{4}} = \sqrt{3}\sqrt{\frac{1}{4}} = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$.

Finally, evaluate $\sqrt{1-u^2}$ at $x = \frac{\sqrt{3}}{2}$ (where $u = \frac{\sqrt{3}}{2}$): $\sqrt{1-u^2} = \sqrt{1-(\frac{\sqrt{3}}{2})^2} = \sqrt{1-\frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

Now, substitute these values into the derivative formula $f'(x) = \frac{-1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$: $f'(\frac{\sqrt{3}}{2}) = \frac{-1}{\frac{1}{2}} \cdot (-2) = (-2) \cdot (-2) = 4$.

Both methods yield the same result.