What is the derivative of y=x^2+4x+3 ?

The derivative of y=x^2+4x+3 is 2x+4

What is the first derivative of y=x^2+4x+3 ?

The first derivative of y=x^2+4x+3 is 2x+4

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derivative of

y = x^{2} + 4 x + 3

2 x + 4

Solution steps

\frac{d}{d x} (x^{2} + 4 x + 3)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{2}) + \frac{d}{d x} (4 x) + \frac{d}{d x} (3)

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

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

\frac{d}{d x} (3) = 0

= 2 x + 4 + 0

Simplify

= 2 x + 4

n = 0 \sum \infty sin (\frac{1}{n ^{2}})

x \to 0 lim (\frac{\frac{1}{1 + x} - 1}{x})

ln (3 - x^{3})

\int ln (x^{2} + x + 1) d x

\int 8 sec^{3} (x) d x

What is the derivative of y=x^2+4x+3 ?
The derivative of y=x^2+4x+3 is 2x+4
What is the first derivative of y=x^2+4x+3 ?
The first derivative of y=x^2+4x+3 is 2x+4