Question

If \( y = \sin^{-1}\!\left(\dfrac{5x + 12\sqrt{1-x^2}}{13}\right) \), then \( \dfrac{dy}{dx} \) is equal to:

• A. \( \dfrac{x}{\sqrt{1-x^2}} \)
• B. \( \dfrac{2}{\sqrt{1-x^2}} \)
• C. \( -\dfrac{1}{\sqrt{1-x^2}} \)
• D. \( -\dfrac{x}{\sqrt{1-x^2}} \)

Hint:

Whenever expressions of the form \[ a\sin\theta + b\cos\theta \] appear, use the identity \[ a\sin\theta + b\cos\theta = \sqrt{a^2+b^2}\sin(\theta+\phi) \] This trick is very useful in inverse trigonometric differentiation problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the derivative of an inverse trigonometric function. The expression inside the \( \sin^{-1} \) is complex, which suggests that a trigonometric substitution is the best approach to simplify it before differentiating.
Step 2: Key Formula or Approach:
The expression \( \frac{5x + 12\sqrt{1-x^2}}{13} \) resembles the trigonometric identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \). Notice that \( 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \). This confirms a Pythagorean triplet, which is a strong hint for this method.
To simplify the expression, we use a substitution. Since the derivative of \( \cos^{-1}(x) \) is \( -1/\sqrt{1-x^2} \), which matches the form of the answer, we should try the substitution \( x = \cos(\theta) \).
Step 3: Detailed Explanation:
1. Perform the substitution:
Let \( x = \cos(\theta) \). This implies \( \theta = \cos^{-1}(x) \).
Then \( \sqrt{1-x^2} = \sqrt{1-\cos^2(\theta)} = \sin(\theta) \).
2. Simplify the expression inside \( \sin^{-1} \):
Substitute \( x \) and \( \sqrt{1-x^2} \) into the expression:
\[ \frac{5x + 12\sqrt{1-x^2}}{13} = \frac{5\cos(\theta) + 12\sin(\theta)}{13} = \frac{5}{13}\cos(\theta) + \frac{12}{13}\sin(\theta) \] Now, let's define a constant angle \( \alpha \) such that \( \sin(\alpha) = \frac{5}{13} \) and \( \cos(\alpha) = \frac{12}{13} \).
The expression becomes:
\[ \sin(\alpha)\cos(\theta) + \cos(\alpha)\sin(\theta) \] This is the expansion of \( \sin(\theta + \alpha) \).
3. Substitute back into the function y:
\[ y = \sin^{-1}(\sin(\theta + \alpha)) \] This simplifies to:
\[ y = \theta + \alpha \] 4. Express y in terms of x and differentiate:
Substitute back \( \theta = \cos^{-1}(x) \).
\[ y = \cos^{-1}(x) + \alpha \] Now, differentiate \( y \) with respect to \( x \). Note that \( \alpha \) is a constant, so its derivative is zero.
\[ \frac{dy}{dx} = \frac{d}{dx}(\cos^{-1}(x) + \alpha) = \frac{d}{dx}(\cos^{-1}(x)) + \frac{d}{dx}(\alpha) \] \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}} + 0 = -\frac{1}{\sqrt{1-x^2}} \] Step 4: Final Answer:
The derivative \( \frac{dy}{dx} \) is \( -\dfrac{1}{\sqrt{1-x^2}} \).