NCSSM Placement Practice Test 2025 – The Comprehensive All-in-One Guide to Succeed!

Question: 1 / 570

What is the derivative of tan inverse(x)?

= 1/(1+x^2)

The derivative of tan inverse(x), also known as the arctangent function, is correctly represented by the expression 1/(1+x^2). This arises from the chain rule in calculus, and it's essential to understand how this particular function behaves.

The arctangent function is the inverse of the tangent function, and it maps the value \(x\) to an angle whose tangent is \(x\). When differentiating this function, we begin by recognizing that the derivative of the tangent function is sec^2(y) when \(y = tan^{-1}(x)\). Specifically, using implicit differentiation, we derive that if \(y = tan^{-1}(x)\), then \(x = tan(y)\).

Differentiating both sides with respect to \(x\), we have:

\[

\frac{dx}{dx} = \frac{d}{dx}[tan(y)]

\]

Utilizing the chain rule, this results in:

\[

1 = sec^2(y) \cdot \frac{dy}{dx}

\]

To express sec^2(y) in terms of \(x\), we utilize the identity that connects secant to tangent:

\[

sec^2(y) = 1 +

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= -1/(1+x^2)

= 1+x^2

= tan(x)/(1+x)

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