What is the derivative of f(x)=x^3-3/2 x^2+4x+1 ?

The derivative of f(x)=x^3-3/2 x^2+4x+1 is 3x^2-3x+4

What is the first derivative of f(x)=x^3-3/2 x^2+4x+1 ?

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

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derivative of

f (x) = x^{3} - \frac{3}{2} x^{2} + 4 x + 1

3 x^{2} - 3 x + 4

Solution steps

\frac{d}{d x} (x^{3} - \frac{3}{2} x^{2} + 4 x + 1)

Simplify

x^{3} - \frac{3}{2} x^{2} + 4 x + 1 : x^{3} - \frac{3 x ^{2}}{2} + 4 x + 1

= \frac{d}{d x} (x^{3} - \frac{3 x ^{2}}{2} + 4 x + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{3}) - \frac{d}{d x} (\frac{3 x ^{2}}{2}) + \frac{d}{d x} (4 x) + \frac{d}{d x} (1)

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

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

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

\frac{d}{d x} (1) = 0

= 3 x^{2} - 3 x + 4 + 0

Simplify

= 3 x^{2} - 3 x + 4

(x - 4) y^{^{'}} + y = x^{2} + 8

3 x^{3}

\frac{d}{d y} (ln (y))

x = - 2 + 4 y, x = y^{2} + 1

\frac{\partial}{\partial x} (10 x cos (4 x y))

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