Question

If \( y = \tan^{-1}(x^2 - x) \), then \( \frac{dy}{dx} = \)

• A. \( \frac{2x}{1+(x^2-x)^2} \)
• B. \( \frac{2x-1}{1+(x^2-x)^2} \)
• C. \( \frac{2x-1}{1-(x^2-x)^2} \)
• D. \( \frac{-2x+1}{1+(x^2-x)^2} \)
• E. \( (2x-1)(1+(x^2-x)^2) \)

Hint:

Always use chain rule with inverse trigonometric functions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires finding the derivative of an inverse tangent function where the argument is itself a function of x. This is a direct application of the chain rule.
Step 2: Key Formula or Approach:
1. Recall the standard derivative for the inverse tangent function: \(\frac{d}{du}(\tan^{-1} u) = \frac{1}{1+u^2}\). 2. Apply the chain rule: If \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). Here, \(f(u) = \tan^{-1} u\) and the inner function is \(g(x) = u = x^2 - x\).
Step 3: Detailed Explanation:
We are given the function \(y = \tan^{-1}(x^2 - x)\). Let the inner function be \(u = x^2 - x\). Then \(y = \tan^{-1}(u)\). Using the chain rule, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\). First, find \(\frac{dy}{du}\): \[ \frac{dy}{du} = \frac{d}{du}(\tan^{-1} u) = \frac{1}{1+u^2} \] Next, find \(\frac{du}{dx}\): \[ \frac{du}{dx} = \frac{d}{dx}(x^2 - x) = 2x - 1 \] Now, multiply the two parts and substitute back \(u = x^2 - x\): \[ \frac{dy}{dx} = \frac{1}{1+u^2} \cdot (2x - 1) = \frac{1}{1+(x^2 - x)^2} \cdot (2x - 1) \] \[ \frac{dy}{dx} = \frac{2x-1}{1+(x^2-x)^2} \] Step 4: Final Answer:
The derivative \(\frac{dy}{dx}\) is \(\frac{2x-1}{1+(x^2-x)^2}\).