What is the derivative of f(x)=x^3-74x^2+40x+27490 ?

The derivative of f(x)=x^3-74x^2+40x+27490 is 3x^2-148x+40

What is the first derivative of f(x)=x^3-74x^2+40x+27490 ?

The first derivative of f(x)=x^3-74x^2+40x+27490 is 3x^2-148x+40

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derivative of

f (x) = x^{3} - 74 x^{2} + 40 x + 27490

3 x^{2} - 148 x + 40

Solution steps

\frac{d}{d x} (x^{3} - 74 x^{2} + 40 x + 27490)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{3}) - \frac{d}{d x} (74 x^{2}) + \frac{d}{d x} (40 x) + \frac{d}{d x} (27490)

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

\frac{d}{d x} (74 x^{2}) = 148 x

\frac{d}{d x} (40 x) = 40

\frac{d}{d x} (27490) = 0

= 3 x^{2} - 148 x + 40 + 0

Simplify

= 3 x^{2} - 148 x + 40

f (x) = sin^{3} (3 x)

f (t) = cos (2 t)

\frac{d}{d x} (2 \cdot cos (x))

\int (2 x^{4} - 8 x^{3}) e^{- 3 x} d x

10 y^{^{''}} + 1000 y = 80 cos (8 t)

What is the derivative of f(x)=x^3-74x^2+40x+27490 ?
The derivative of f(x)=x^3-74x^2+40x+27490 is 3x^2-148x+40
What is the first derivative of f(x)=x^3-74x^2+40x+27490 ?
The first derivative of f(x)=x^3-74x^2+40x+27490 is 3x^2-148x+40