If y = tan(2x + 1), dy/dx = ?

If y = tan(2x + 1), dy/dx = ?

• A. sec^2(2x+1)
• B. 2 sec^2(2x+1)
• C. 2 tan^2(2x+1)
• D. 4 sec^2(2x+1)

Answer: (B)

When differentiating a composition like tan(u) with u = 2x + 1, use the chain rule. The derivative of tan(u) with respect to u is sec^2(u), and then you multiply by du/dx to account for the inner function. Here, du/dx = d/dx(2x + 1) = 2. So the derivative is sec^2(2x + 1) multiplied by 2, giving 2 sec^2(2x + 1). This matches the idea that the slope of tan at an angle is sec^2 of that angle, and since the inner function changes twice as fast as x, the slope doubles. The other forms miss the chain-rule factor or replace sec^2 with tan^2, which does not match the derivative of tangent.

When differentiating a composition like tan(u) with u = 2x + 1, use the chain rule. The derivative of tan(u) with respect to u is sec^2(u), and then you multiply by du/dx to account for the inner function.

Here, du/dx = d/dx(2x + 1) = 2. So the derivative is sec^2(2x + 1) multiplied by 2, giving 2 sec^2(2x + 1).

This matches the idea that the slope of tan at an angle is sec^2 of that angle, and since the inner function changes twice as fast as x, the slope doubles. The other forms miss the chain-rule factor or replace sec^2 with tan^2, which does not match the derivative of tangent.