What is the derivative of f(x)=(x^3+1)*3x^2 ?

The derivative of f(x)=(x^3+1)*3x^2 is 3(5x^4+2x)

What is the first derivative of f(x)=(x^3+1)*3x^2 ?

The first derivative of f(x)=(x^3+1)*3x^2 is 3(5x^4+2x)

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derivative of

f (x) = (x^{3} + 1) \cdot 3 x^{2}

3 (5 x^{4} + 2 x)

Solution steps

\frac{d}{d x} ((x^{3} + 1) \cdot 3 x^{2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d x} ((x^{3} + 1) x^{2})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{3} + 1, g = x^{2}

= 3 (\frac{d}{d x} (x^{3} + 1) x^{2} + \frac{d}{d x} (x^{2}) (x^{3} + 1))

\frac{d}{d x} (x^{3} + 1) = 3 x^{2}

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

= 3 (3 x^{2} x^{2} + 2 x (x^{3} + 1))

Simplify

3 (3 x^{2} x^{2} + 2 x (x^{3} + 1)) : 3 (5 x^{4} + 2 x)

= 3 (5 x^{4} + 2 x)

t sin (9 t)

y = x^{2} - x, y = 0

\frac{2 s + 9}{s ^{2} + 4 s + 40}

\frac{\partial}{\partial t} (\frac{t}{c ^{2} t ^{2} - x ^{2}})

y^{^{''}} + 12 y^{^{'}} + 36 = 0

What is the derivative of f(x)=(x^3+1)*3x^2 ?
The derivative of f(x)=(x^3+1)*3x^2 is 3(5x^4+2x)
What is the first derivative of f(x)=(x^3+1)*3x^2 ?
The first derivative of f(x)=(x^3+1)*3x^2 is 3(5x^4+2x)