What is the derivative of (3x^4+1/(x^3)) ?

The derivative of (3x^4+1/(x^3)) is (3(x^4-1))/(x^4)

What is the first derivative of (3x^4+1/(x^3)) ?

The first derivative of (3x^4+1/(x^3)) is (3(x^4-1))/(x^4)

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\frac{d}{d x} (\frac{3 x ^{4} + 1}{x ^{3}})

\frac{3 ( x ^{4} - 1 )}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{3 x ^{4} + 1}{x ^{3}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( 3 x ^{4} + 1 ) x ^{3} - \frac{d}{d x} ( x ^{3} ) ( 3 x ^{4} + 1 )}{( x ^{3} ) ^{2}}

\frac{d}{d x} (3 x^{4} + 1) = 12 x^{3}

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

= \frac{12 x ^{3} x ^{3} - 3 x ^{2} ( 3 x ^{4} + 1 )}{( x ^{3} ) ^{2}}

Simplify

\frac{12 x ^{3} x ^{3} - 3 x ^{2} ( 3 x ^{4} + 1 )}{( x ^{3} ) ^{2}} : \frac{3 ( x ^{4} - 1 )}{x ^{4}}

= \frac{3 ( x ^{4} - 1 )}{x ^{4}}

\frac{d}{d x} (2 - \frac{x ^{2} - 2 x + 4}{x - 2})

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

x^{2} \frac{d ^{2} y}{d x ^{2}} - 5 x y^{^{'}} + 8 y = 4 x^{2}

\int_{0}^{1} 2 π (x + 1) (x^{2}) d x

y = (3)^{x}, at x = 6

What is the derivative of (3x^4+1/(x^3)) ?
The derivative of (3x^4+1/(x^3)) is (3(x^4-1))/(x^4)
What is the first derivative of (3x^4+1/(x^3)) ?
The first derivative of (3x^4+1/(x^3)) is (3(x^4-1))/(x^4)