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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 2, y (1) = 3

x \to 7 + lim (ln (x - 7))

\int e^{7 x + 9} d x

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

e^{4 x} cos (7 x)

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