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Given that c(q) = (q + 1)^2 - q, find dc/dq.

Question:

Given that {eq}c(q) = (q + 1)^2 - q {/eq}, find {eq}\frac{\mathrm{d} c}{\mathrm{d} q} {/eq}.

Chain Rule:

The chain rule is useful for calculating derivatives of composite functions. It is given by the formula {eq}f(g(x))' = f'(g(x))g'(x) {/eq}

Answer and Explanation:

Taking the derivative:

{eq}c'(q) = \left( (q + 1)^2 - q \right)' \\ c'(q) = \left( (q + 1)^2 \right)' - \left( q \right)' \\ c'(q) = \left( (q + 1)^2 \right)' - 1 {/eq}

Using the chain rule:

{eq}c'(q) = 2 (q + 1) \left( q+1 \right)' - 1 \\ c'(q) = 2 (q + 1) \left( 1 \right) - 1 \\ c'(q) = 2 (q + 1) - 1 \\ c'(q) = 2q + 1 {/eq}

Answer and Explanation:

Taking the derivative:

{eq}c'(q) = \left( (q + 1)^2 - q \right)' \\ c'(q) = \left( (q + 1)^2 \right)' - \left( q \right)' \\ c'(q) = \left( (q + 1)^2 \right)' - 1 {/eq}

Using the chain rule:

{eq}c'(q) = 2 (q + 1) \left( q+1 \right)' - 1 \\ c'(q) = 2 (q + 1) \left( 1 \right) - 1 \\ c'(q) = 2 (q + 1) - 1 \\ c'(q) = 2q + 1 {/eq}


Learn more about this topic:

Using the Chain Rule to Differentiate Complex Functions

from Math 104: Calculus

Chapter 7 / Lesson 6
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