What is the derivative of (sqrt(x)-5/(sqrt(x)+6)) ?

The derivative of (sqrt(x)-5/(sqrt(x)+6)) is (11)/(2sqrt(x)(\sqrt{x)+6)^2}

What is the first derivative of (sqrt(x)-5/(sqrt(x)+6)) ?

The first derivative of (sqrt(x)-5/(sqrt(x)+6)) is (11)/(2sqrt(x)(\sqrt{x)+6)^2}

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\frac{d}{d x} (\frac{x - 5}{x + 6})

\frac{11}{2 x ( x + 6 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x - 5}{x + 6})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x - 5 ) ( x + 6 ) - \frac{d}{d x} ( x + 6 ) ( x - 5 )}{( x + 6 ) ^{2}}

\frac{d}{d x} (x - 5) = \frac{1}{2 x}

\frac{d}{d x} (x + 6) = \frac{1}{2 x}

= \frac{\frac{1}{2 x} ( x + 6 ) - \frac{1}{2 x} ( x - 5 )}{( x + 6 ) ^{2}}

Simplify

\frac{\frac{1}{2 x} ( x + 6 ) - \frac{1}{2 x} ( x - 5 )}{( x + 6 ) ^{2}} : \frac{11}{2 x ( x + 6 ) ^{2}}

= \frac{11}{2 x ( x + 6 ) ^{2}}

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

y^{^{'}} = t + 3

\frac{\partial}{\partial x} (\frac{( 3 x + 5 )}{6 y + 1})

\int_{0}^{2} π (4 x^{2} - x^{4}) d x

y = cos (x), at x = \frac{π}{4}

What is the derivative of (sqrt(x)-5/(sqrt(x)+6)) ?
The derivative of (sqrt(x)-5/(sqrt(x)+6)) is (11)/(2sqrt(x)(\sqrt{x)+6)^2}
What is the first derivative of (sqrt(x)-5/(sqrt(x)+6)) ?
The first derivative of (sqrt(x)-5/(sqrt(x)+6)) is (11)/(2sqrt(x)(\sqrt{x)+6)^2}