What is the derivative of 0.002x^2+1.7x+60 ?

The derivative of 0.002x^2+1.7x+60 is 0.004x+1.7

What is the first derivative of 0.002x^2+1.7x+60 ?

The first derivative of 0.002x^2+1.7x+60 is 0.004x+1.7

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derivative of

0.002 x^{2} + 1.7 x + 60

0.004 x + 1.7

Solution steps

\frac{d}{d x} (0.002 x^{2} + 1.7 x + 60)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (0.002 x^{2}) + \frac{d}{d x} (1.7 x) + \frac{d}{d x} (60)

\frac{d}{d x} (0.002 x^{2}) = 0.004 x

\frac{d}{d x} (1.7 x) = 1.7

\frac{d}{d x} (60) = 0

= 0.004 x + 1.7 + 0

Simplify

= 0.004 x + 1.7

x \to 0 lim (co tan (x) - \frac{1}{x})

G (t) = \frac{1 - 2 t}{3 + t}

(1.2) (3.5)

\frac{d}{d x} ((x + csc (x^{3} + 3))^{- 3})

y = \frac{e ^{x}}{8 x ^{4}}

What is the derivative of 0.002x^2+1.7x+60 ?
The derivative of 0.002x^2+1.7x+60 is 0.004x+1.7
What is the first derivative of 0.002x^2+1.7x+60 ?
The first derivative of 0.002x^2+1.7x+60 is 0.004x+1.7