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

The derivative of ln((x^3(2x+1))/(x^2-3)) is (4x^3+x^2-24x-9)/(x(2x+1)(x^2-3))

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

The first derivative of ln((x^3(2x+1))/(x^2-3)) is (4x^3+x^2-24x-9)/(x(2x+1)(x^2-3))

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derivative of

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

\frac{4 x ^{3} + x ^{2} - 24 x - 9}{x ( 2 x + 1 ) ( x ^{2} - 3 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{3} ( 2 x + 1 )}{x ^{2} - 3}) = \frac{4 x ^{5} + x ^{4} - 24 x ^{3} - 9 x ^{2}}{( x ^{2} - 3 ) ^{2}}

= \frac{1}{\frac{x ^{3} ( 2 x + 1 )}{x ^{2} - 3}} \cdot \frac{4 x ^{5} + x ^{4} - 24 x ^{3} - 9 x ^{2}}{( x ^{2} - 3 ) ^{2}}

Simplify

\frac{1}{\frac{x ^{3} ( 2 x + 1 )}{x ^{2} - 3}} \cdot \frac{4 x ^{5} + x ^{4} - 24 x ^{3} - 9 x ^{2}}{( x ^{2} - 3 ) ^{2}} : \frac{4 x ^{3} + x ^{2} - 24 x - 9}{x ( 2 x + 1 ) ( x ^{2} - 3 )}

= \frac{4 x ^{3} + x ^{2} - 24 x - 9}{x ( 2 x + 1 ) ( x ^{2} - 3 )}

\frac{\partial}{\partial t} (cos (t) - sin (t))

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

\frac{\partial}{\partial x} (\frac{x + y}{1 + x ^{2}})

\int \frac{e ^{4 x}}{1 + e ^{4 x}} d x

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the derivative of ln((x^3(2x+1))/(x^2-3)) ?
The derivative of ln((x^3(2x+1))/(x^2-3)) is (4x^3+x^2-24x-9)/(x(2x+1)(x^2-3))
What is the first derivative of ln((x^3(2x+1))/(x^2-3)) ?
The first derivative of ln((x^3(2x+1))/(x^2-3)) is (4x^3+x^2-24x-9)/(x(2x+1)(x^2-3))