Question:
If \( y = \tan^{-1}(x^2 - x) \), then \( \frac{dy}{dx} = \)
If \( y = \tan^{-1}(x^2 - x) \), then \( \frac{dy}{dx} = \)
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Always use chain rule with inverse trigonometric functions.
Updated On: Apr 21, 2026
- \( \frac{2x}{1+(x^2-x)^2} \)
- \( \frac{2x-1}{1+(x^2-x)^2} \)
- \( \frac{2x-1}{1-(x^2-x)^2} \)
- \( \frac{-2x+1}{1+(x^2-x)^2} \)
- \( (2x-1)(1+(x^2-x)^2) \)
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The Correct Option is B
Solution and Explanation
Concept:
\[
\frac{d}{dx}(\tan^{-1}u) = \frac{u'}{1+u^2}
\]
Step 1: Differentiate inner function.
\[ u = x^2 - x \Rightarrow u' = 2x - 1 \]
Step 2: Apply formula.
\[ \frac{dy}{dx} = \frac{2x-1}{1+(x^2-x)^2} \]
Step 1: Differentiate inner function.
\[ u = x^2 - x \Rightarrow u' = 2x - 1 \]
Step 2: Apply formula.
\[ \frac{dy}{dx} = \frac{2x-1}{1+(x^2-x)^2} \]
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