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

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

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

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

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derivative of

y = x^{3} + 4

3 x^{2}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (4) = 0

= 3 x^{2} + 0

Simplify

= 3 x^{2}

\frac{d}{d x} (cos^{2} (sin (5 e^{x} - 1 6^{x})))

\int \frac{1}{t ^{3}} d t

\frac{\partial}{\partial x} ((x) cos (x) cos (y))

y^{^{'}} = \frac{sec ( y )}{( x - 4 ) ^{2}}

(3, - 4), (5, 6)

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