What is the first d/(dq)(-q^2+36) ?

The first d/(dq)(-q^2+36) is -2q

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d/(dq)(-q^2+36)

\frac{d}{d q} (- q^{2} + 36)

- 2 q

Solution steps

\frac{d}{d q} (- q^{2} + 36)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d q} (q^{2}) + \frac{d}{d q} (36)

\frac{d}{d q} (q^{2}) = 2 q

\frac{d}{d q} (36) = 0

= - 2 q + 0

Simplify

= - 2 q

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