What is the derivative of 21x^3+9/x ?

The derivative of 21x^3+9/x is 63x^2-9/(x^2)

What is the first derivative of 21x^3+9/x ?

The first derivative of 21x^3+9/x is 63x^2-9/(x^2)

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\frac{d}{d x} (21 x^{3} + \frac{9}{x})

63 x^{2} - \frac{9}{x ^{2}}

Solution steps

\frac{d}{d x} (21 x^{3} + \frac{9}{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (21 x^{3}) + \frac{d}{d x} (\frac{9}{x})

\frac{d}{d x} (21 x^{3}) = 63 x^{2}

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

= 63 x^{2} - \frac{9}{x ^{2}}

\frac{\partial}{\partial x} (x e^{\frac{x}{y}})

(6, - 12), (10, - 8)

\int \frac{x ^{2} + 2 x}{x ^{3} + 3 x ^{2} + 1} d x

\int \frac{( ( ln ( x ) ) ^{3} )}{x} d x

What is the derivative of 21x^3+9/x ?
The derivative of 21x^3+9/x is 63x^2-9/(x^2)
What is the first derivative of 21x^3+9/x ?
The first derivative of 21x^3+9/x is 63x^2-9/(x^2)