In Class 12 Calculus, you must memorize the derivatives of all six inverse trigonometric functions. The derivative of $\tan^{-1}(x)$—also written as $\arctan(x)$—is one of the most frequently used formulas in both integration and differential equations.
Notice that the derivative of $\tan^{-1}(x)$ contains no square roots or negative signs, unlike the derivatives for inverse sine or cosine.
Because integration is the reverse process, $\int \frac{1}{1 + x^2} dx = \tan^{-1}(x) + C$.
The derivative of $\tan^{-1}(x)$ with respect to $x$ is:
$\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1 + x^2}$
To prove why this formula is true, we use implicit differentiation and basic trigonometry.
Step 1: Let $y = \tan^{-1}(x)$. Therefore, we can rewrite this as: $\tan(y) = x$
Step 2: Differentiate both sides with respect to $x$. The derivative of $\tan(y)$ requires the chain rule: $\sec^2(y) \cdot \frac{dy}{dx}$. The derivative of $x$ is just $1$. So: $\sec^2(y) \cdot \frac{dy}{dx} = 1$
Step 3: Isolate $\frac{dy}{dx}$. $\frac{dy}{dx} = \frac{1}{\sec^2(y)}$
Step 4: Convert 'y' back to 'x' using a trig identity. We know the fundamental identity: $\sec^2(y) = 1 + \tan^2(y)$. Substitute this into our fraction: $\frac{dy}{dx} = \frac{1}{1 + \tan^2(y)}$
Since we established in Step 1 that $\tan(y) = x$, then $\tan^2(y)$ must be $x^2$. Substitute $x^2$ in: $\frac{dy}{dx} = \frac{1}{1 + x^2}$. (Proof complete).
The derivative of cot^-1(x) is exactly the same as tan^-1(x), but with a negative sign: -1 / (1 + x²).
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