Question

Multiple Choice Which of the following is \(\frac{d}{d x} \tan ^{-1}(3 x) ?\) \((\mathbf{A})-\frac{3}{1+9 x^{2}} \quad(\mathbf{B})-\frac{1}{1+9 x^{2}} \quad\) (C) \(\frac{1}{1+9 x^{2}}\) \((\mathbf{D}) \frac{3}{1+9 x^{2}} \quad(\mathbf{E}) \frac{3}{\sqrt{1-9 x^{2}}}\)

Short answer

Therefore, the derivative of \(\tan^{-1}(3x)\) is \(\frac{3}{1+9x^2}\). So, the correct option is (D). \(\frac{3}{1+9x^2}\)

Step-by-step solution

Recall the derivative of the inverse tangent

The derivative of the inverse tangent can be expressed as \( \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}\).

Apply the chain rule

To calculate the derivative of \(\tan^{-1}(3x)\), the chain rule is applied. In this case, the outer function is \(\tan^{-1}(x)\) and the inner function is \(3x\). Apply the rule to calculate the derivative as \(\frac{1}{1+(3x)^2} * \frac{d}{dx}(3x)\), which simplifies to \(\frac{3}{1+9x^2}\).