What is the (-x^3+4x^2-5x)^' ?

The (-x^3+4x^2-5x)^' is -3x^2+8x-5

What is the first (-x^3+4x^2-5x)^' ?

The first (-x^3+4x^2-5x)^' is -3x^2+8x-5

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

- 3 x^{2} + 8 x - 5

Solution steps

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

Apply the Sum/Difference Rule:

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

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

(x^{3})^{'} = 3 x^{2}

(4 x^{2})^{'} = 8 x

(5 x)^{'} = 5

= - 3 x^{2} + 8 x - 5

(x, y) \to (π, \frac{π}{2}) lim (y sin (x - y))

f (x) = (x^{2} - 8)^{5}, at x = 3

x \to 5.6 lim (7.3 x + 8.4)

\int_{0}^{\frac{1}{2}} arccos (x) d x

(x, y) \to (0, 0) lim (\frac{y}{x})