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

The derivative of ln(((x+1(x+2))/(x+3))) is (x^2+6x+7)/((x+1)(x+2)(x+3))

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

The first derivative of ln(((x+1(x+2))/(x+3))) is (x^2+6x+7)/((x+1)(x+2)(x+3))

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{( x + 1 ) ( x + 2 )}{x + 3}) = \frac{x ^{2} + 6 x + 7}{( x + 3 ) ^{2}}

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

Simplify

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

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

\int_{- 1}^{1} x^{2} d x

\int 4 e^{x + y}

\int cosh (u) d u

y = x^{2} - 5, (- 5, 20)

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

What is the derivative of ln(((x+1(x+2))/(x+3))) ?
The derivative of ln(((x+1(x+2))/(x+3))) is (x^2+6x+7)/((x+1)(x+2)(x+3))
What is the first derivative of ln(((x+1(x+2))/(x+3))) ?
The first derivative of ln(((x+1(x+2))/(x+3))) is (x^2+6x+7)/((x+1)(x+2)(x+3))