What is the first d/(du)(u^2-v^2) ?

The first d/(du)(u^2-v^2) is 2u

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d/(du)(u^2-v^2)

\frac{d}{d u} (u^{2} - v^{2})

2 u

Solution steps

\frac{d}{d u} (u^{2} - v^{2})

Treat

v

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (v^{2})

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

\frac{d}{d u} (v^{2}) = 0

= 2 u - 0

Simplify

= 2 u

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