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

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

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

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

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derivative of

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

2 x + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

= 2 x + 1 - 0

Simplify

= 2 x + 1

\frac{d y}{d x} = e^{6 y + 6 x}

\frac{s + 5}{s ( s ^{2} + 8 s + 25 )}

\frac{1}{3 x ^{\frac{2}{3}}} + 2

\int 3 x (x^{2} + 6)^{2} d x

\frac{\partial}{\partial y} (5 cos (x y^{2}))

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