What is the derivative of f(x)=(x^2+1)/(x^3+1) ?

The derivative of f(x)=(x^2+1)/(x^3+1) is (-x^4-3x^2+2x)/((x^3+1)^2)

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

The first derivative of f(x)=(x^2+1)/(x^3+1) is (-x^4-3x^2+2x)/((x^3+1)^2)

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Expand

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

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

f (x) = \frac{x + 6}{x + 1}

\frac{d}{d x} (20 t - \frac{1}{2} g (t, x) t^{2})

lo g_{e} (x)

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

\int_{0}^{\frac{1}{2}} x ln (2 x) d x

What is the derivative of f(x)=(x^2+1)/(x^3+1) ?
The derivative of f(x)=(x^2+1)/(x^3+1) is (-x^4-3x^2+2x)/((x^3+1)^2)
What is the first derivative of f(x)=(x^2+1)/(x^3+1) ?
The first derivative of f(x)=(x^2+1)/(x^3+1) is (-x^4-3x^2+2x)/((x^3+1)^2)