What is the derivative of {z}(x+x^2) ?

The derivative of {z}(x+x^2) is derivative of {z}(x)+2x

What is the first derivative of {z}(x+x^2) ?

The first derivative of {z}(x+x^2) is derivative of {z}(x)+2x

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

\frac{d}{d x} (z (x)) + 2 x

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

= \frac{d}{d x} (z (x)) + 2 x

\int_{0}^{1} t 1 + t^{2} + t^{4} d t

x \to 3 - lim (sin (π x) + 1)

\int e^{2 x + e^{x}} d x

x \to 1 + lim (\frac{x + 10}{x + 2})

2 x y \frac{d y}{d x} = y^{2} - x^{2}

What is the derivative of {z}(x+x^2) ?
The derivative of {z}(x+x^2) is derivative of {z}(x)+2x
What is the first derivative of {z}(x+x^2) ?
The first derivative of {z}(x+x^2) is derivative of {z}(x)+2x