Question

determine \\(\frac{dy}{dx}\\) for \\(y = \tan^{-1}(11x + 3)\\).

Explanation:

Identify the outer and inner functions

We are given the function:

$$ y = \tan^{-1}(11x + 3) $$

To find the derivative \(\frac{dy}{dx}\), we identify the outer function as \(f(u) = \tan^{-1}(u)\) and the inner function as \(u(x) = 11x + 3\).

Differentiate the outer function

The derivative of the inverse tangent function is:

$$ \frac{d}{du}[\tan^{-1}(u)] = \frac{1}{1 + u^2} $$

Apply the chain rule

Using the Chain Rule Differentiation knowledge point

$$ \frac{dy}{dx} = \frac{d}{du}[\tan^{-1}(u)] \cdot \frac{du}{dx} = \frac{1}{1 + u^2} \cdot \frac{d}{dx}[11x + 3] $$

Compute the derivative of the inner function

We find the derivative of the inner linear function:

$$ \frac{d}{dx}[11x + 3] = 11 $$

Substitute and simplify the expression

Substituting \(u = 11x + 3\) and the inner derivative back into the chain rule formula:

$$ \frac{dy}{dx} = \frac{1}{1 + (11x + 3)^2} \cdot 11 = \frac{11}{1 + (11x + 3)^2} $$

Answer:

\(\frac{dy}{dx} = \frac{11}{1 + (11x + 3)^2}\)