What is the derivative of x^3+4x-15 ?

The derivative of x^3+4x-15 is 3x^2+4

What is the first derivative of x^3+4x-15 ?

The first derivative of x^3+4x-15 is 3x^2+4

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\frac{d}{d x} (x^{3} + 4 x - 15)

3 x^{2} + 4

Solution steps

\frac{d}{d x} (x^{3} + 4 x - 15)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{3}) + \frac{d}{d x} (4 x) - \frac{d}{d x} (15)

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

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

\frac{d}{d x} (15) = 0

= 3 x^{2} + 4 - 0

Simplify

= 3 x^{2} + 4

f (x) = (3 x - 2)^{5} (x + 1)^{3}

y - 2 y^{^{'}} + y = 0

\frac{d}{d x} (x^{4 c o s (x)})

k = 4 \sum \infty 3 (- \frac{1}{2})^{k}

What is the derivative of x^3+4x-15 ?
The derivative of x^3+4x-15 is 3x^2+4
What is the first derivative of x^3+4x-15 ?
The first derivative of x^3+4x-15 is 3x^2+4