What is the derivative of ((x^5))/5-((16x^3))/3+7 ?

The derivative of ((x^5))/5-((16x^3))/3+7 is x^4-16x^2

What is the first derivative of ((x^5))/5-((16x^3))/3+7 ?

The first derivative of ((x^5))/5-((16x^3))/3+7 is x^4-16x^2

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derivative of

\frac{( x ^{5} )}{5} - \frac{( 16 x ^{3} )}{3} + 7

x^{4} - 16 x^{2}

Solution steps

\frac{d}{d x} (\frac{x ^{5}}{5} - \frac{16 x ^{3}}{3} + 7)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{5}}{5}) - \frac{d}{d x} (\frac{16 x ^{3}}{3}) + \frac{d}{d x} (7)

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

\frac{d}{d x} (\frac{16 x ^{3}}{3}) = 16 x^{2}

\frac{d}{d x} (7) = 0

= x^{4} - 16 x^{2} + 0

Simplify

= x^{4} - 16 x^{2}

\frac{d}{d t} (4 + 3 t + 2 t^{2})

p + 1

- 3 cos^{3} (x)

\frac{1}{x ^{2} + a ^{2}}

3 e^{2} x

What is the derivative of ((x^5))/5-((16x^3))/3+7 ?
The derivative of ((x^5))/5-((16x^3))/3+7 is x^4-16x^2
What is the first derivative of ((x^5))/5-((16x^3))/3+7 ?
The first derivative of ((x^5))/5-((16x^3))/3+7 is x^4-16x^2