What is the derivative of (sqrt(x+1)-1/x) ?

The derivative of (sqrt(x+1)-1/x) is (-x-2+2sqrt(x+1))/(2x^2sqrt(x+1))

What is the first derivative of (sqrt(x+1)-1/x) ?

The first derivative of (sqrt(x+1)-1/x) is (-x-2+2sqrt(x+1))/(2x^2sqrt(x+1))

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\frac{d}{d x} (\frac{x + 1 - 1}{x})

\frac{- x - 2 + 2 x + 1}{2 x ^{2} x + 1}

Solution steps

\frac{d}{d x} (\frac{x + 1 - 1}{x})

Apply the Quotient Rule:

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

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

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

\frac{d x}{d x} = 1

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

Simplify

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

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

\frac{d}{d x} (x (24 - 2 x) (24 - 2 x))

x \to - 2 lim (\frac{5 x + 2}{x - 2})

\int \frac{1}{0.02 y - 6} d y

f (x) = x^{- 1}

\int_{0}^{2 π} \frac{1 - cos ( x )}{2} d x

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