What is the derivative of (x^4/3-7x) ?

The derivative of (x^4/3-7x) is (4x^3)/3-7

What is the first derivative of (x^4/3-7x) ?

The first derivative of (x^4/3-7x) is (4x^3)/3-7

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\frac{d}{d x} (\frac{x ^{4}}{3} - 7 x)

\frac{4 x ^{3}}{3} - 7

Solution steps

\frac{d}{d x} (\frac{x ^{4}}{3} - 7 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{4}}{3}) - \frac{d}{d x} (7 x)

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

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

= \frac{4 x ^{3}}{3} - 7

e^{- 2 t}

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

n = 1 \sum \infty \frac{( - 1 )}{n}

\frac{s}{( s ^{2} + 3 ^{2} ) ^{2}}

n = 2 \sum \infty \frac{1}{4 ^{n}}

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