What is the derivative of f(x)=ln(x^3+2x^2+2x) ?

The derivative of f(x)=ln(x^3+2x^2+2x) is (3x^2+4x+2)/(x^3+2x^2+2x)

What is the first derivative of f(x)=ln(x^3+2x^2+2x) ?

The first derivative of f(x)=ln(x^3+2x^2+2x) is (3x^2+4x+2)/(x^3+2x^2+2x)

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derivative of

f (x) = ln (x^{3} + 2 x^{2} + 2 x)

\frac{3 x ^{2} + 4 x + 2}{x ^{3} + 2 x ^{2} + 2 x}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{3 x ^{2} + 4 x + 2}{x ^{3} + 2 x ^{2} + 2 x}

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

\int \frac{( 2 x + 1 )}{x ^{2} - 2 x + 5} d x

x \to \frac{π}{4} lim (tan^{2} (x))

\int 9 tan (x) d x

x \to - 1 lim (\frac{x + 2}{x + 1})

What is the derivative of f(x)=ln(x^3+2x^2+2x) ?
The derivative of f(x)=ln(x^3+2x^2+2x) is (3x^2+4x+2)/(x^3+2x^2+2x)
What is the first derivative of f(x)=ln(x^3+2x^2+2x) ?
The first derivative of f(x)=ln(x^3+2x^2+2x) is (3x^2+4x+2)/(x^3+2x^2+2x)