What is the first derivative of 3x+y^2 ?

The first derivative of 3x+y^2 is 3

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derivative of 3x+y^2

derivative of

3 x + y^{2}

3

Solution steps

\frac{d}{d x} (3 x + y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (3 x) + \frac{d}{d x} (y^{2})

\frac{d}{d x} (3 x) = 3

\frac{d}{d x} (y^{2}) = 0

= 3 + 0

Simplify

= 3

tan (e^{s}) in x^{2}

(1 + x)^{2 y + 1}

y = (sin (x))^{5}

(2^{x})^{2} + 1 + e^{c} os x

g (x) = 2 x + 2