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

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

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

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

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derivative of

y = x^{2} + 4 x

2 x + 4

Solution steps

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

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} (x^{2}) = 2 x

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

= 2 x + 4

n = 0 \sum \infty 0.9 9^{n}

\int 3 z^{2} d z

x \to 0 lim ((e^{x})^{2})

\int \frac{7 x ^{3} - 3}{x ^{2} - 4} d x

\frac{d}{d x} (x^{3} - 7 x + 6)

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