Question

The derivative of \( t^2 + t \) with respect to \( t-1 \) at \( t = -2 \), is equal to:

• A. -4
• B. 2
• C. -1
• D. -3
• E. \( -\frac{1}{2} \)

Hint:

When calculating derivatives with respect to a different variable, remember to apply the chain rule and carefully substitute the values to avoid errors.

Answer 1

The Correct Option is D

Step 1: To find the derivative of \( t^2 + t \) with respect to \( t-1 \), first use the chain rule: \[ \frac{d}{d(t-1)} \left(t^2 + t\right) = \frac{d}{dt} \left(t^2 + t\right) \times \frac{dt}{d(t-1)}. \] Step 2: Now, calculate the derivative of \( t^2 + t \): \[ \frac{d}{dt} (t^2 + t) = 2t + 1. \] Step 3: Since \( \frac{dt}{d(t-1)} = 1 \), the derivative is simply \( 2t + 1 \). 
Step 4: Substitute \( t = -2 \) into the derivative: \[ 2(-2) + 1 = -4 + 1 = -3. \] Therefore, the derivative of \( t^2 + t \) with respect to \( t-1 \) at \( t = -2 \) is \( -3 \).