What is the derivative of 2x^2+x-1 ?

The derivative of 2x^2+x-1 is 4x+1

What is the first derivative of 2x^2+x-1 ?

The first derivative of 2x^2+x-1 is 4x+1

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derivative of

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

x \to - 6 lim (\frac{( x ^{2} - 36 )}{x + 6})

\int \frac{2 e ^{x}}{e ^{2 x} + 1} d x

(4 x^{2} + 3 y^{2}) d x - 2 x y d y = 0

\int_{1}^{5} x^{3} ln (4 x) d x

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

What is the derivative of 2x^2+x-1 ?
The derivative of 2x^2+x-1 is 4x+1
What is the first derivative of 2x^2+x-1 ?
The first derivative of 2x^2+x-1 is 4x+1