What is the derivative of ax^2+bx ?

The derivative of ax^2+bx is 2ax+b

What is the first derivative of ax^2+bx ?

The first derivative of ax^2+bx is 2ax+b

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\frac{d}{d x} (a x^{2} + b x)

2 a x + b

Solution steps

\frac{d}{d x} (a x^{2} + b x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (a x^{2}) + \frac{d}{d x} (b x)

\frac{d}{d x} (a x^{2}) = 2 a x

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

= 2 a x + b

\int_{1}^{2} (\frac{1}{8} + \frac{3}{8} x) d x

f (x) = \frac{cos ( x )}{x ^{4}}

x \to - 2 - lim (\frac{x - 3}{x ^{2} - 4})

\int sin (a x) cos (a) x d x

\frac{d}{d x} (x^{2} arctan (5 x) + ln (1 + 4 x^{2}))

What is the derivative of ax^2+bx ?
The derivative of ax^2+bx is 2ax+b
What is the first derivative of ax^2+bx ?
The first derivative of ax^2+bx is 2ax+b