What is the derivative of f(x)=x^2+1.8x-0.1404 ?

The derivative of f(x)=x^2+1.8x-0.1404 is 2x+1.8

What is the first derivative of f(x)=x^2+1.8x-0.1404 ?

The first derivative of f(x)=x^2+1.8x-0.1404 is 2x+1.8

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derivative of

f (x) = x^{2} + 1.8 x - 0.1404

2 x + 1.8

Solution steps

\frac{d}{d x} (x^{2} + 1.8 x - 0.1404)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{2}) + \frac{d}{d x} (1.8 x) - \frac{d}{d x} (0.1404)

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

\frac{d}{d x} (1.8 x) = 1.8

\frac{d}{d x} (0.1404) = 0

= 2 x + 1.8 - 0

Simplify

= 2 x + 1.8

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What is the derivative of f(x)=x^2+1.8x-0.1404 ?
The derivative of f(x)=x^2+1.8x-0.1404 is 2x+1.8
What is the first derivative of f(x)=x^2+1.8x-0.1404 ?
The first derivative of f(x)=x^2+1.8x-0.1404 is 2x+1.8