What is the derivative of 1+40x^3-3x^5 ?

The derivative of 1+40x^3-3x^5 is -15x^4+120x^2

What is the first derivative of 1+40x^3-3x^5 ?

The first derivative of 1+40x^3-3x^5 is -15x^4+120x^2

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\frac{d}{d x} (1 + 40 x^{3} - 3 x^{5})

- 15 x^{4} + 120 x^{2}

Solution steps

\frac{d}{d x} (1 + 40 x^{3} - 3 x^{5})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (1) + \frac{d}{d x} (40 x^{3}) - \frac{d}{d x} (3 x^{5})

\frac{d}{d x} (1) = 0

\frac{d}{d x} (40 x^{3}) = 120 x^{2}

\frac{d}{d x} (3 x^{5}) = 15 x^{4}

= 0 + 120 x^{2} - 15 x^{4}

Simplify

= - 15 x^{4} + 120 x^{2}

x \to 2 lim (1680 x^{3} - 120 x)

\int \frac{2 x - 5}{x ^{2} + 2 x + 2} d x

\int \frac{1}{8 x} d x

n = 0 \sum \infty \frac{1}{n} x^{n}

\int sec^{3} (z) d z

What is the derivative of 1+40x^3-3x^5 ?
The derivative of 1+40x^3-3x^5 is -15x^4+120x^2
What is the first derivative of 1+40x^3-3x^5 ?
The first derivative of 1+40x^3-3x^5 is -15x^4+120x^2