What is the first (40q^2+q+500)^' ?

The first (40q^2+q+500)^' is 80q+1

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(40q^2+q+500)^'

(40 q^{2} + q + 500)^{^{'}}

80 q + 1

Solution steps

(40 q^{2} + q + 500)^{'}

Apply the Sum/Difference Rule:

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

= (40 q^{2})^{^{'}} + q^{^{'}} + 50 0^{^{'}}

(40 q^{2})^{'} = 80 q

q^{'} = 1

50 0^{'} = 0

= 80 q + 1 + 0

Simplify

= 80 q + 1

x \to - 1 + lim (\frac{3}{x + 1})

y = e^{x} + 1

x \to 0 lim (e^{- x} x^{2})

\int - 2 x^{2} + 4 x + 6 d x

\int_{0}^{1} e^{- 0.1 x} d x