What is the derivative of (x^5-8x^3+x+6/x) ?

The derivative of (x^5-8x^3+x+6/x) is (4x^5-16x^3-6)/(x^2)

What is the first derivative of (x^5-8x^3+x+6/x) ?

The first derivative of (x^5-8x^3+x+6/x) is (4x^5-16x^3-6)/(x^2)

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\frac{d}{d x} (\frac{x ^{5} - 8 x ^{3} + x + 6}{x})

\frac{4 x ^{5} - 16 x ^{3} - 6}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{5} - 8 x ^{3} + x + 6}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{5} - 8 x ^{3} + x + 6 ) x - \frac{d x}{d x} ( x ^{5} - 8 x ^{3} + x + 6 )}{x ^{2}}

\frac{d}{d x} (x^{5} - 8 x^{3} + x + 6) = 5 x^{4} - 24 x^{2} + 1

\frac{d x}{d x} = 1

= \frac{( 5 x ^{4} - 24 x ^{2} + 1 ) x - 1 \cdot ( x ^{5} - 8 x ^{3} + x + 6 )}{x ^{2}}

Simplify

\frac{( 5 x ^{4} - 24 x ^{2} + 1 ) x - 1 \cdot ( x ^{5} - 8 x ^{3} + x + 6 )}{x ^{2}} : \frac{4 x ^{5} - 16 x ^{3} - 6}{x ^{2}}

= \frac{4 x ^{5} - 16 x ^{3} - 6}{x ^{2}}

x \to 0 lim (2 x^{3} - 4 x + 2)

\int (\frac{2}{x} + 3 sin (x) - 4 e^{x}) d x

y = 1 + 3 x, 1 \leq x \leq 9

x \to \infty lim (a^{x})

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

What is the derivative of (x^5-8x^3+x+6/x) ?
The derivative of (x^5-8x^3+x+6/x) is (4x^5-16x^3-6)/(x^2)
What is the first derivative of (x^5-8x^3+x+6/x) ?
The first derivative of (x^5-8x^3+x+6/x) is (4x^5-16x^3-6)/(x^2)