What is the derivative of f(x)=ax^2+bx ?

The derivative of f(x)=ax^2+bx is 2ax+b

What is the first derivative of f(x)=ax^2+bx ?

The first derivative of f(x)=ax^2+bx is 2ax+b

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derivative of

f (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 \frac{2}{x ^{0.85}} d x

\frac{\partial}{\partial x} (2 x y + x f (x, y) (\frac{y}{x}))

\frac{\partial}{\partial y} (3 + 2 x y)

5 x, 3 x^{2}

\frac{\partial}{\partial x} (2 cos (x^{2} + y^{2}))

What is the derivative of f(x)=ax^2+bx ?
The derivative of f(x)=ax^2+bx is 2ax+b
What is the first derivative of f(x)=ax^2+bx ?
The first derivative of f(x)=ax^2+bx is 2ax+b