What is the derivative of ln^4(((x^2-1^3)/(x^2-1))) ?

The derivative of ln^4(((x^2-1^3)/(x^2-1))) is (128xln^3(x^2-1))/(x^2-1)

What is the first derivative of ln^4(((x^2-1^3)/(x^2-1))) ?

The first derivative of ln^4(((x^2-1^3)/(x^2-1))) is (128xln^3(x^2-1))/(x^2-1)

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

\frac{128 x ln ^{3} ( x ^{2} - 1 )}{x ^{2} - 1}

Solution steps

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

Simplify

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

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

Apply the chain rule:

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

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

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

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

Simplify

4 ln^{3} ((x^{2} - 1)^{2}) \frac{4 x}{x ^{2} - 1} : \frac{128 x ln ^{3} ( x ^{2} - 1 )}{x ^{2} - 1}

= \frac{128 x ln ^{3} ( x ^{2} - 1 )}{x ^{2} - 1}

\int sin (x) sin (cos (x)) d x

f (x) = 5 - x^{2}, at x = - 1

\int \frac{1}{x ^{6} - x ^{3}} d x

\int (4 x^{3} + cos (x)) d x

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}

What is the derivative of ln^4(((x^2-1^3)/(x^2-1))) ?
The derivative of ln^4(((x^2-1^3)/(x^2-1))) is (128xln^3(x^2-1))/(x^2-1)
What is the first derivative of ln^4(((x^2-1^3)/(x^2-1))) ?
The first derivative of ln^4(((x^2-1^3)/(x^2-1))) is (128xln^3(x^2-1))/(x^2-1)