What is the derivative of f(x)= x/(x+b/x) ?

The derivative of f(x)= x/(x+b/x) is (2bx)/((x^2+b)^2)

What is the first derivative of f(x)= x/(x+b/x) ?

The first derivative of f(x)= x/(x+b/x) is (2bx)/((x^2+b)^2)

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derivative of

f (x) = \frac{x}{x + \frac{b}{x}}

\frac{2 b x}{( x ^{2} + b ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{2 b x}{( x ^{2} + b ) ^{2}}

f (x) = - 2 cos (x), at x = - π

\frac{\partial}{\partial y} (2 x yz)

\int \frac{sec ( x ) tan ( x )}{tan ( x )} d x

x + 1

f (x) = 2 x - 1

What is the derivative of f(x)= x/(x+b/x) ?
The derivative of f(x)= x/(x+b/x) is (2bx)/((x^2+b)^2)
What is the first derivative of f(x)= x/(x+b/x) ?
The first derivative of f(x)= x/(x+b/x) is (2bx)/((x^2+b)^2)