What is the derivative of x^2sec(1/x) ?

The derivative of x^2sec(1/x) is 2xsec(1/x)-sec(1/x)tan(1/x)

What is the first derivative of x^2sec(1/x) ?

The first derivative of x^2sec(1/x) is 2xsec(1/x)-sec(1/x)tan(1/x)

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\frac{d}{d x} (x^{2} sec (\frac{1}{x}))

2 x sec (\frac{1}{x}) - sec (\frac{1}{x}) tan (\frac{1}{x})

Solution steps

\frac{d}{d x} (x^{2} sec (\frac{1}{x}))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = sec (\frac{1}{x})

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

\frac{d}{d x} (x^{2}) = 2 x

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

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

Simplify

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

= 2 x sec (\frac{1}{x}) - sec (\frac{1}{x}) tan (\frac{1}{x})

\int x^{3.4} - 2 x^{2 - 1} d x

\int_{- 2}^{0} 4 - x^{2} d x

\int x^{3} sin (x) d x

\frac{\partial}{\partial t} (t^{4} e^{- x})

\frac{2}{1 - 2 x}

What is the derivative of x^2sec(1/x) ?
The derivative of x^2sec(1/x) is 2xsec(1/x)-sec(1/x)tan(1/x)
What is the first derivative of x^2sec(1/x) ?
The first derivative of x^2sec(1/x) is 2xsec(1/x)-sec(1/x)tan(1/x)