What is the derivative of (2x^6+1)/(8x^2) ?

The derivative of (2x^6+1)/(8x^2) is (4x^6-1)/(4x^3)

What is the first derivative of (2x^6+1)/(8x^2) ?

The first derivative of (2x^6+1)/(8x^2) is (4x^6-1)/(4x^3)

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derivative of

\frac{2 x ^{6} + 1}{8 x ^{2}}

\frac{4 x ^{6} - 1}{4 x ^{3}}

Solution steps

\frac{d}{d x} (\frac{2 x ^{6} + 1}{8 x ^{2}})

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (\frac{2 x ^{6} + 1}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{1}{8} \cdot \frac{\frac{d}{d x} ( 2 x ^{6} + 1 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( 2 x ^{6} + 1 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (2 x^{6} + 1) = 12 x^{5}

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

= \frac{1}{8} \cdot \frac{12 x ^{5} x ^{2} - 2 x ( 2 x ^{6} + 1 )}{( x ^{2} ) ^{2}}

Simplify

\frac{1}{8} \cdot \frac{12 x ^{5} x ^{2} - 2 x ( 2 x ^{6} + 1 )}{( x ^{2} ) ^{2}} : \frac{4 x ^{6} - 1}{4 x ^{3}}

= \frac{4 x ^{6} - 1}{4 x ^{3}}

\int \frac{cos ( u )}{3 a} d u

\int_{0}^{1} 12 cos (x^{2}) d x

x \to \frac{π}{4} lim (cot (4 x))

n = 0 \sum \infty sin (2 n)

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

What is the derivative of (2x^6+1)/(8x^2) ?
The derivative of (2x^6+1)/(8x^2) is (4x^6-1)/(4x^3)
What is the first derivative of (2x^6+1)/(8x^2) ?
The first derivative of (2x^6+1)/(8x^2) is (4x^6-1)/(4x^3)