What is the derivative of f(x)=0.0622x^2+1.1119x+2.3979 ?

The derivative of f(x)=0.0622x^2+1.1119x+2.3979 is 0.1244x+1.1119

What is the first derivative of f(x)=0.0622x^2+1.1119x+2.3979 ?

The first derivative of f(x)=0.0622x^2+1.1119x+2.3979 is 0.1244x+1.1119

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derivative of

f (x) = 0.0622 x^{2} + 1.1119 x + 2.3979

0.1244 x + 1.1119

Solution steps

\frac{d}{d x} (0.0622 x^{2} + 1.1119 x + 2.3979)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (0.0622 x^{2}) + \frac{d}{d x} (1.1119 x) + \frac{d}{d x} (2.3979)

\frac{d}{d x} (0.0622 x^{2}) = 0.1244 x

\frac{d}{d x} (1.1119 x) = 1.1119

\frac{d}{d x} (2.3979) = 0

= 0.1244 x + 1.1119 + 0

Simplify

= 0.1244 x + 1.1119

y = x^{5}

\int \frac{10}{x} d x

R (q) = (q^{2} + 1)^{4}

x \to 3 + lim (\frac{4 x}{x - 3})

y = 8 (x - \frac{1}{x})^{4}, at x = 2

What is the derivative of f(x)=0.0622x^2+1.1119x+2.3979 ?
The derivative of f(x)=0.0622x^2+1.1119x+2.3979 is 0.1244x+1.1119
What is the first derivative of f(x)=0.0622x^2+1.1119x+2.3979 ?
The first derivative of f(x)=0.0622x^2+1.1119x+2.3979 is 0.1244x+1.1119