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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\frac{d}{d x} (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 )}

f (x) = 6 cos^{4} (3 x + 2)

x \to 5 lim (x^{3} - x^{2} + 5 x)

y^{^{'}} = y tan (t) + sin (t)

\frac{d}{d x} ((\frac{1}{x} - \frac{4}{x ^{3}}) (x^{3} - 5 x - 1))

x \to 1 lim (x^{5} - 2 x - 6)

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))