What is the derivative of f(x)= 1/2 b(1/x-1/(x^0))^2 ?

The derivative of f(x)= 1/2 b(1/x-1/(x^0))^2 is -(b(1-x))/(x^3)

What is the first derivative of f(x)= 1/2 b(1/x-1/(x^0))^2 ?

The first derivative of f(x)= 1/2 b(1/x-1/(x^0))^2 is -(b(1-x))/(x^3)

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derivative of

f (x) = \frac{1}{2} b (\frac{1}{x} - \frac{1}{x ^{0}})^{2}

- \frac{b ( 1 - x )}{x ^{3}}

Solution steps

\frac{d}{d x} (\frac{1}{2} b (\frac{1}{x} - \frac{1}{x ^{0}})^{2})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} b \frac{d}{d x} ((\frac{1}{x} - \frac{1}{x ^{0}})^{2})

Apply the chain rule:

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

= 2 (\frac{1}{x} - \frac{1}{x ^{0}}) \frac{d}{d x} (\frac{1}{x} - \frac{1}{x ^{0}})

\frac{d}{d x} (\frac{1}{x} - \frac{1}{x ^{0}}) = - \frac{1}{x ^{2}}

= \frac{1}{2} b \cdot 2 (\frac{1}{x} - \frac{1}{x ^{0}}) (- \frac{1}{x ^{2}})

Simplify

\frac{1}{2} b \cdot 2 (\frac{1}{x} - \frac{1}{x ^{0}}) (- \frac{1}{x ^{2}}) : - \frac{b ( 1 - x )}{x ^{3}}

= - \frac{b ( 1 - x )}{x ^{3}}

f (x) = x^{3} - 5 x^{2} + 7, at x = 1

\int \frac{x + 5}{x ^{2} + 2 x - 15} d x

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{3}} d x

x \to 5 + lim (x^{2} - 25)

\int 0.6 x^{2} d x

What is the derivative of f(x)= 1/2 b(1/x-1/(x^0))^2 ?
The derivative of f(x)= 1/2 b(1/x-1/(x^0))^2 is -(b(1-x))/(x^3)
What is the first derivative of f(x)= 1/2 b(1/x-1/(x^0))^2 ?
The first derivative of f(x)= 1/2 b(1/x-1/(x^0))^2 is -(b(1-x))/(x^3)