What is the derivative of (12x^6-3x^2+x^2/(3x^4)) ?

The derivative of (12x^6-3x^2+x^2/(3x^4)) is (4(6x^4+1))/(3x^3)

What is the first derivative of (12x^6-3x^2+x^2/(3x^4)) ?

The first derivative of (12x^6-3x^2+x^2/(3x^4)) is (4(6x^4+1))/(3x^3)

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\frac{d}{d x} (\frac{12 x ^{6} - 3 x ^{2} + x ^{2}}{3 x ^{4}})

\frac{4 ( 6 x ^{4} + 1 )}{3 x ^{3}}

Solution steps

\frac{d}{d x} (\frac{12 x ^{6} - 3 x ^{2} + x ^{2}}{3 x ^{4}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{3} \frac{d}{d x} (\frac{12 x ^{6} - 3 x ^{2} + x ^{2}}{x ^{4}})

Apply the Quotient Rule:

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

= \frac{1}{3} \cdot \frac{\frac{d}{d x} ( 12 x ^{6} - 3 x ^{2} + x ^{2} ) x ^{4} - \frac{d}{d x} ( x ^{4} ) ( 12 x ^{6} - 3 x ^{2} + x ^{2} )}{( x ^{4} ) ^{2}}

\frac{d}{d x} (12 x^{6} - 3 x^{2} + x^{2}) = 72 x^{5} - 4 x

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

= \frac{1}{3} \cdot \frac{( 72 x ^{5} - 4 x ) x ^{4} - 4 x ^{3} ( 12 x ^{6} - 3 x ^{2} + x ^{2} )}{( x ^{4} ) ^{2}}

Simplify

\frac{1}{3} \cdot \frac{( 72 x ^{5} - 4 x ) x ^{4} - 4 x ^{3} ( 12 x ^{6} - 3 x ^{2} + x ^{2} )}{( x ^{4} ) ^{2}} : \frac{4 ( 6 x ^{4} + 1 )}{3 x ^{3}}

= \frac{4 ( 6 x ^{4} + 1 )}{3 x ^{3}}

\int cos (x) \cdot sin^{2} (x) d x

\int x^{\frac{5}{3}} d x

\int sec^{2} (8 x) d x

(y \cdot e^{- x}) \frac{d y}{d x} = x

\int_{0}^{π} 1 + cos (2 x) d x

What is the derivative of (12x^6-3x^2+x^2/(3x^4)) ?
The derivative of (12x^6-3x^2+x^2/(3x^4)) is (4(6x^4+1))/(3x^3)
What is the first derivative of (12x^6-3x^2+x^2/(3x^4)) ?
The first derivative of (12x^6-3x^2+x^2/(3x^4)) is (4(6x^4+1))/(3x^3)