What is the derivative of 40x^4+180x^3+240x^2+100x ?

The derivative of 40x^4+180x^3+240x^2+100x is 160x^3+540x^2+480x+100

What is the first derivative of 40x^4+180x^3+240x^2+100x ?

The first derivative of 40x^4+180x^3+240x^2+100x is 160x^3+540x^2+480x+100

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\frac{d}{d x} (40 x^{4} + 180 x^{3} + 240 x^{2} + 100 x)

160 x^{3} + 540 x^{2} + 480 x + 100

Solution steps

\frac{d}{d x} (40 x^{4} + 180 x^{3} + 240 x^{2} + 100 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (40 x^{4}) + \frac{d}{d x} (180 x^{3}) + \frac{d}{d x} (240 x^{2}) + \frac{d}{d x} (100 x)

\frac{d}{d x} (40 x^{4}) = 160 x^{3}

\frac{d}{d x} (180 x^{3}) = 540 x^{2}

\frac{d}{d x} (240 x^{2}) = 480 x

\frac{d}{d x} (100 x) = 100

= 160 x^{3} + 540 x^{2} + 480 x + 100

\frac{d}{d x} (\frac{1}{3} (3 - 5 x)^{3})

\int_{0}^{2} \frac{16}{21 x ^{2}} d x

9 x^{2}, at x = 2

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

\frac{\partial}{\partial x} (\frac{a}{x ^{2}})

What is the derivative of 40x^4+180x^3+240x^2+100x ?
The derivative of 40x^4+180x^3+240x^2+100x is 160x^3+540x^2+480x+100
What is the first derivative of 40x^4+180x^3+240x^2+100x ?
The first derivative of 40x^4+180x^3+240x^2+100x is 160x^3+540x^2+480x+100