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

The derivative of ln((x-1^3(x^2+1)^4)) is (11x^2-8x+3)/((x-1)(x^2+1))

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

The first derivative of ln((x-1^3(x^2+1)^4)) is (11x^2-8x+3)/((x-1)(x^2+1))

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

\frac{11 x ^{2} - 8 x + 3}{( x - 1 ) ( x ^{2} + 1 )}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{( x - 1 ) ^{3} ( x ^{2} + 1 ) ^{4}} (3 (x - 1)^{2} (x^{2} + 1)^{4} + 8 x (x^{2} + 1)^{3} (x - 1)^{3}) : \frac{11 x ^{2} - 8 x + 3}{( x - 1 ) ( x ^{2} + 1 )}

= \frac{11 x ^{2} - 8 x + 3}{( x - 1 ) ( x ^{2} + 1 )}

\frac{d}{d x} (3 x^{2} - e^{2 x})

\frac{d y}{d t} = - \frac{3 y}{t} - 2 - t^{- 4}

\frac{\partial}{\partial z} (cos (x y) + sin (yz))

\int sin^{3} (x) cos^{2} (x) d x

\int_{1}^{e^{2}} \frac{ln ^{5} ( x ^{2} )}{x} d x

What is the derivative of ln((x-1^3(x^2+1)^4)) ?
The derivative of ln((x-1^3(x^2+1)^4)) is (11x^2-8x+3)/((x-1)(x^2+1))
What is the first derivative of ln((x-1^3(x^2+1)^4)) ?
The first derivative of ln((x-1^3(x^2+1)^4)) is (11x^2-8x+3)/((x-1)(x^2+1))