What is the derivative of f(x)=2x^2+x-1 ?

The derivative of f(x)=2x^2+x-1 is 4x+1

What is the first derivative of f(x)=2x^2+x-1 ?

The first derivative of f(x)=2x^2+x-1 is 4x+1

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derivative of

f (x) = 2 x^{2} + x - 1

4 x + 1

Solution steps

\frac{d}{d x} (2 x^{2} + x - 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 x^{2}) + \frac{d x}{d x} - \frac{d}{d x} (1)

\frac{d}{d x} (2 x^{2}) = 4 x

\frac{d x}{d x} = 1

\frac{d}{d x} (1) = 0

= 4 x + 1 - 0

Simplify

= 4 x + 1

n = 4 \sum \infty n! (x - 4)^{n}

3 x e^{2 x}

t^{2} \frac{d y}{d t} + t y = 3

\frac{d}{d t} (5 t + \frac{7}{t})

ln (x), 1, - 1

What is the derivative of f(x)=2x^2+x-1 ?
The derivative of f(x)=2x^2+x-1 is 4x+1
What is the first derivative of f(x)=2x^2+x-1 ?
The first derivative of f(x)=2x^2+x-1 is 4x+1