What is the first derivative of kx+d ?

The first derivative of kx+d is k

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derivative of kx+d

\frac{d}{d x} (k x + d)

k

Solution steps

\frac{d}{d x} (k x + d)

Treat

k, d

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (k x) + \frac{dd}{d x}

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

\frac{dd}{d x} = 0

= k + 0

Simplify

= k

\frac{d}{d x} (2 (x - 2)^{2} + 1)

\int \frac{1}{2 x ^{3} - 3 x ^{2}} d x

x^{2} (3 - x)

c y^{- 6}

x \to 1 lim (x^{3} - 2 x^{2} + x - 3)