What is the first derivative of t^2+5 ?

The first derivative of t^2+5 is 2t

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derivative of t^2+5

derivative of

t^{2} + 5

2 t

Solution steps

\frac{d}{d t} (t^{2} + 5)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{2}) + \frac{d}{d t} (5)

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

\frac{d}{d t} (5) = 0

= 2 t + 0

Simplify

= 2 t

\frac{d}{d x} (\frac{1}{( x ^{5} - x + 1 ) ^{9}})

y^{^{''}} - 26 y^{^{'}} + 169 y = 0

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