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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d y}{d t} = y^{3}, y (0) = 1

\frac{d}{d x} (\frac{432}{x})

\int \frac{4 x ^{2} - 7 x - 12}{x ( x + 2 ) ( x - 3 )} d x

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

x^{3}, 3

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