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

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

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

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

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\frac{d}{d x} (x^{2} + x - 4)

2 x + 1

Solution steps

\frac{d}{d x} (x^{2} + x - 4)

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} (4)

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

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

\frac{d}{d x} (4) = 0

= 2 x + 1 - 0

Simplify

= 2 x + 1

\frac{d}{d x} (x^{2} - 3 x)

\int_{1}^{5} 3 x^{2} + \frac{1}{12 x ^{2}} d x

t = y + y^{^{'}}

\int 5 - x^{2} d x

\int_{- 3}^{3} (2 x^{4} - 3 x^{3}) d x

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