What is the derivative of 2/5 x^5-3x^2+8x-5 ?

The derivative of 2/5 x^5-3x^2+8x-5 is 2x^4-6x+8

What is the first derivative of 2/5 x^5-3x^2+8x-5 ?

The first derivative of 2/5 x^5-3x^2+8x-5 is 2x^4-6x+8

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\frac{d}{d x} (\frac{2}{5} x^{5} - 3 x^{2} + 8 x - 5)

2 x^{4} - 6 x + 8

Solution steps

\frac{d}{d x} (\frac{2}{5} x^{5} - 3 x^{2} + 8 x - 5)

Simplify

\frac{2}{5} x^{5} - 3 x^{2} + 8 x - 5 : \frac{2 x ^{5}}{5} - 3 x^{2} + 8 x - 5

= \frac{d}{d x} (\frac{2 x ^{5}}{5} - 3 x^{2} + 8 x - 5)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{2 x ^{5}}{5}) - \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (8 x) - \frac{d}{d x} (5)

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

\frac{d}{d x} (3 x^{2}) = 6 x

\frac{d}{d x} (8 x) = 8

\frac{d}{d x} (5) = 0

= 2 x^{4} - 6 x + 8 - 0

Simplify

= 2 x^{4} - 6 x + 8

y = csc^{2} (x) - 2

5 y^{^{''}} + y^{^{'}} = 0

y^{^{'}} = sin (3 x)

\frac{d}{d x} (ln (cos (0)))

\int \frac{4 x + 6}{x ^{2} + 4} d x

What is the derivative of 2/5 x^5-3x^2+8x-5 ?
The derivative of 2/5 x^5-3x^2+8x-5 is 2x^4-6x+8
What is the first derivative of 2/5 x^5-3x^2+8x-5 ?
The first derivative of 2/5 x^5-3x^2+8x-5 is 2x^4-6x+8