What is the derivative of 0.03x^2-4x+3x^{0.8}-5000 ?

The derivative of 0.03x^2-4x+3x^{0.8}-5000 is 0.06x+(2.4)/(x^{0.2)}-4

What is the first derivative of 0.03x^2-4x+3x^{0.8}-5000 ?

The first derivative of 0.03x^2-4x+3x^{0.8}-5000 is 0.06x+(2.4)/(x^{0.2)}-4

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derivative of

0.03 x^{2} - 4 x + 3 x^{0.8} - 5000

0.06 x + \frac{2.4}{x ^{0.2}} - 4

Solution steps

\frac{d}{d x} (0.03 x^{2} - 4 x + 3 x^{0.8} - 5000)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (0.03 x^{2}) - \frac{d}{d x} (4 x) + \frac{d}{d x} (3 x^{0.8}) - \frac{d}{d x} (5000)

\frac{d}{d x} (0.03 x^{2}) = 0.06 x

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

\frac{d}{d x} (3 x^{0.8}) = \frac{2.4}{x ^{0.2}}

\frac{d}{d x} (5000) = 0

= 0.06 x - 4 + \frac{2.4}{x ^{0.2}} - 0

Simplify

= 0.06 x + \frac{2.4}{x ^{0.2}} - 4

\int - \frac{t}{1 + t ^{2}} d t

f (x) = 14 cos (x), at x = 7

y^{^{'}} + ln (x) = \frac{y}{x}

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

\frac{( 6 )}{s ^{2} + 5 s + 6}

What is the derivative of 0.03x^2-4x+3x^{0.8}-5000 ?
The derivative of 0.03x^2-4x+3x^{0.8}-5000 is 0.06x+(2.4)/(x^{0.2)}-4
What is the first derivative of 0.03x^2-4x+3x^{0.8}-5000 ?
The first derivative of 0.03x^2-4x+3x^{0.8}-5000 is 0.06x+(2.4)/(x^{0.2)}-4