What is the derivative of-0.02x^3+0.37x^2+241x+17905 ?

The derivative of-0.02x^3+0.37x^2+241x+17905 is -0.06x^2+0.74x+241

What is the first derivative of-0.02x^3+0.37x^2+241x+17905 ?

The first derivative of-0.02x^3+0.37x^2+241x+17905 is -0.06x^2+0.74x+241

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derivative of

- 0.02 x^{3} + 0.37 x^{2} + 241 x + 17905

- 0.06 x^{2} + 0.74 x + 241

Solution steps

\frac{d}{d x} (- 0.02 x^{3} + 0.37 x^{2} + 241 x + 17905)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (0.02 x^{3}) + \frac{d}{d x} (0.37 x^{2}) + \frac{d}{d x} (241 x) + \frac{d}{d x} (17905)

\frac{d}{d x} (0.02 x^{3}) = 0.06 x^{2}

\frac{d}{d x} (0.37 x^{2}) = 0.74 x

\frac{d}{d x} (241 x) = 241

\frac{d}{d x} (17905) = 0

= - 0.06 x^{2} + 0.74 x + 241 + 0

Simplify

= - 0.06 x^{2} + 0.74 x + 241

(\frac{t}{t + 1 ( t )})^{^{'}}

y^{^{'}} + \frac{y}{x} = \frac{1}{x ^{2}}

y = 4 sin (π x + y), (1, 0)

n = 0 \sum \infty 6 (- \frac{2}{3})^{n}

y^{^{'}} = \frac{y ^{4} - 3}{( 4 x y ^{3} )}

What is the derivative of-0.02x^3+0.37x^2+241x+17905 ?
The derivative of-0.02x^3+0.37x^2+241x+17905 is -0.06x^2+0.74x+241
What is the first derivative of-0.02x^3+0.37x^2+241x+17905 ?
The first derivative of-0.02x^3+0.37x^2+241x+17905 is -0.06x^2+0.74x+241