What is the first derivative of s^2+9 ?

The first derivative of s^2+9 is 2s

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derivative of s^2+9

derivative of

s^{2} + 9

2 s

Solution steps

\frac{d}{d s} (s^{2} + 9)

Apply the Sum/Difference Rule:

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

= \frac{d}{d s} (s^{2}) + \frac{d}{d s} (9)

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

\frac{d}{d s} (9) = 0

= 2 s + 0

Simplify

= 2 s

4 x^{3} - 2 x^{2}

\frac{d}{d r} (r^{3} - r^{2})

\frac{d}{d x} (x_{6} x^{2} + 9)

x = 5 t^{2} - 2 t + 7

\frac{d}{d a} (1 - 8 sin^{2} (a) cos^{2} (a))