What is the (ln((x(x^2+1)^2)/(sqrt(2x^3-1))))^' ?

The (ln((x(x^2+1)^2)/(sqrt(2x^3-1))))^' is (7x^7+6x^5-5x^4-x^3-6x^2-1)/(x(x^2+1)^2(2x^3-1))

What is the first (ln((x(x^2+1)^2)/(sqrt(2x^3-1))))^' ?

The first (ln((x(x^2+1)^2)/(sqrt(2x^3-1))))^' is (7x^7+6x^5-5x^4-x^3-6x^2-1)/(x(x^2+1)^2(2x^3-1))

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(ln (\frac{x ( x ^{2} + 1 ) ^{2}}{2 x ^{3} - 1}))^{^{'}}

\frac{7 x ^{7} + 6 x ^{5} - 5 x ^{4} - x ^{3} - 6 x ^{2} - 1}{x ( x ^{2} + 1 ) ^{2} ( 2 x ^{3} - 1 )}

Solution steps

(ln (\frac{x ( x ^{2} + 1 ) ^{2}}{2 x ^{3} - 1}))^{'}

Apply the chain rule:

\frac{1}{\frac{x ( x ^{2} + 1 ) ^{2}}{2 x ^{3} - 1}} (\frac{x ( x ^{2} + 1 ) ^{2}}{2 x ^{3} - 1})^{'}

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

(\frac{x ( x ^{2} + 1 ) ^{2}}{2 x ^{3} - 1})^{'} = \frac{7 x ^{7} + 6 x ^{5} - 5 x ^{4} - x ^{3} - 6 x ^{2} - 1}{( 2 x ^{3} - 1 ) 2 x ^{3} - 1}

= \frac{1}{\frac{x ( x ^{2} + 1 ) ^{2}}{2 x ^{3} - 1}} \cdot \frac{7 x ^{7} + 6 x ^{5} - 5 x ^{4} - x ^{3} - 6 x ^{2} - 1}{( 2 x ^{3} - 1 ) 2 x ^{3} - 1}

Simplify

\frac{1}{\frac{x ( x ^{2} + 1 ) ^{2}}{2 x ^{3} - 1}} \cdot \frac{7 x ^{7} + 6 x ^{5} - 5 x ^{4} - x ^{3} - 6 x ^{2} - 1}{( 2 x ^{3} - 1 ) 2 x ^{3} - 1} : \frac{7 x ^{7} + 6 x ^{5} - 5 x ^{4} - x ^{3} - 6 x ^{2} - 1}{x ( x ^{2} + 1 ) ^{2} ( 2 x ^{3} - 1 )}

= \frac{7 x ^{7} + 6 x ^{5} - 5 x ^{4} - x ^{3} - 6 x ^{2} - 1}{x ( x ^{2} + 1 ) ^{2} ( 2 x ^{3} - 1 )}

f (x) = 1 + sin (x), at x = - π

\int x^{3} ln (8 x) d x

x \to - 3 lim (3)

\int \frac{5 ( 4 x + 2 )}{2 ( x ^{2} + 4 x + 5 )} d x

f (x) = \frac{x ^{2}}{( 1 + x )}

What is the (ln((x(x^2+1)^2)/(sqrt(2x^3-1))))^' ?
The (ln((x(x^2+1)^2)/(sqrt(2x^3-1))))^' is (7x^7+6x^5-5x^4-x^3-6x^2-1)/(x(x^2+1)^2(2x^3-1))
What is the first (ln((x(x^2+1)^2)/(sqrt(2x^3-1))))^' ?
The first (ln((x(x^2+1)^2)/(sqrt(2x^3-1))))^' is (7x^7+6x^5-5x^4-x^3-6x^2-1)/(x(x^2+1)^2(2x^3-1))