What is the derivative of f(x)=-3/2 x^3+2/5 x^2-4 ?

The derivative of f(x)=-3/2 x^3+2/5 x^2-4 is -(9x^2)/2+(4x)/5

What is the first derivative of f(x)=-3/2 x^3+2/5 x^2-4 ?

The first derivative of f(x)=-3/2 x^3+2/5 x^2-4 is -(9x^2)/2+(4x)/5

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derivative of

f (x) = - \frac{3}{2} x^{3} + \frac{2}{5} x^{2} - 4

- \frac{9 x ^{2}}{2} + \frac{4 x}{5}

Solution steps

\frac{d}{d x} (- \frac{3}{2} x^{3} + \frac{2}{5} x^{2} - 4)

Simplify

- \frac{3}{2} x^{3} + \frac{2}{5} x^{2} - 4 : - \frac{3 x ^{3}}{2} + \frac{2 x ^{2}}{5} - 4

= \frac{d}{d x} (- \frac{3 x ^{3}}{2} + \frac{2 x ^{2}}{5} - 4)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (\frac{3 x ^{3}}{2}) + \frac{d}{d x} (\frac{2 x ^{2}}{5}) - \frac{d}{d x} (4)

\frac{d}{d x} (\frac{3 x ^{3}}{2}) = \frac{9 x ^{2}}{2}

\frac{d}{d x} (\frac{2 x ^{2}}{5}) = \frac{4 x}{5}

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

= - \frac{9 x ^{2}}{2} + \frac{4 x}{5} - 0

Simplify

= - \frac{9 x ^{2}}{2} + \frac{4 x}{5}

g (t) = \frac{7}{8 t ^{5}}

\int 3 sin (6 x) d x

\frac{d ^{2}}{d x ^{2}} (- \frac{x}{y})

y^{^{''}} + 36 y = 6 t^{2}

5^{x^{3}}

What is the derivative of f(x)=-3/2 x^3+2/5 x^2-4 ?
The derivative of f(x)=-3/2 x^3+2/5 x^2-4 is -(9x^2)/2+(4x)/5
What is the first derivative of f(x)=-3/2 x^3+2/5 x^2-4 ?
The first derivative of f(x)=-3/2 x^3+2/5 x^2-4 is -(9x^2)/2+(4x)/5