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

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

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

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

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derivative of

y = \frac{x ^{4}}{4} + \frac{2}{9} x^{3} - x^{2} + 9 x - 5

x^{3} + \frac{2 x ^{2}}{3} - 2 x + 9

Solution steps

\frac{d}{d x} (\frac{x ^{4}}{4} + \frac{2}{9} x^{3} - x^{2} + 9 x - 5)

Simplify

\frac{x ^{4}}{4} + \frac{2}{9} x^{3} - x^{2} + 9 x - 5 : \frac{x ^{4}}{4} + \frac{2 x ^{3}}{9} - x^{2} + 9 x - 5

= \frac{d}{d x} (\frac{x ^{4}}{4} + \frac{2 x ^{3}}{9} - x^{2} + 9 x - 5)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{4}}{4}) + \frac{d}{d x} (\frac{2 x ^{3}}{9}) - \frac{d}{d x} (x^{2}) + \frac{d}{d x} (9 x) - \frac{d}{d x} (5)

\frac{d}{d x} (\frac{x ^{4}}{4}) = x^{3}

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

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

\frac{d}{d x} (9 x) = 9

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

= x^{3} + \frac{2 x ^{2}}{3} - 2 x + 9 - 0

Simplify

= x^{3} + \frac{2 x ^{2}}{3} - 2 x + 9

\frac{( 2 x )}{1 + x ^{4}}

7 - 4 x

\frac{d}{d x} (e^{4 x s i n (2 x)})

cot^{2} (x)

f (x) = ln (196 sin^{2} (x))

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