What is the derivative of (sqrt(x)-7/(sqrt(x)+1)) ?

The derivative of (sqrt(x)-7/(sqrt(x)+1)) is 4/(sqrt(x)(\sqrt{x)+1)^2}

What is the first derivative of (sqrt(x)-7/(sqrt(x)+1)) ?

The first derivative of (sqrt(x)-7/(sqrt(x)+1)) is 4/(sqrt(x)(\sqrt{x)+1)^2}

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\frac{d}{d x} (\frac{x - 7}{x + 1})

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

Solution steps

\frac{d}{d x} (\frac{x - 7}{x + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x - 7 ) ( x + 1 ) - \frac{d}{d x} ( x + 1 ) ( x - 7 )}{( x + 1 ) ^{2}}

\frac{d}{d x} (x - 7) = \frac{1}{2 x}

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

= \frac{\frac{1}{2 x} ( x + 1 ) - \frac{1}{2 x} ( x - 7 )}{( x + 1 ) ^{2}}

Simplify

\frac{\frac{1}{2 x} ( x + 1 ) - \frac{1}{2 x} ( x - 7 )}{( x + 1 ) ^{2}} : \frac{4}{x ( x + 1 ) ^{2}}

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

\int 8 cos^{3} (x) d x

\frac{\partial}{\partial x} (p (x))

\frac{d}{d x} (x^{2} sin^{4} (x) + x cos^{- 2} (x))

y^{^{'}} - 6 y = 2 e^{t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = - 4, y^{^{'}} (0) = - 6

What is the derivative of (sqrt(x)-7/(sqrt(x)+1)) ?
The derivative of (sqrt(x)-7/(sqrt(x)+1)) is 4/(sqrt(x)(\sqrt{x)+1)^2}
What is the first derivative of (sqrt(x)-7/(sqrt(x)+1)) ?
The first derivative of (sqrt(x)-7/(sqrt(x)+1)) is 4/(sqrt(x)(\sqrt{x)+1)^2}