What is the derivative of x^2sec(1/(x^3)) ?

The derivative of x^2sec(1/(x^3)) is 2xsec(1/(x^3))-(3sec(1/(x^3))tan(1/(x^3)))/(x^2)

What is the first derivative of x^2sec(1/(x^3)) ?

The first derivative of x^2sec(1/(x^3)) is 2xsec(1/(x^3))-(3sec(1/(x^3))tan(1/(x^3)))/(x^2)

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

2 x sec (\frac{1}{x ^{3}}) - \frac{3 sec ( \frac{1}{x ^{3}} ) tan ( \frac{1}{x ^{3}} )}{x ^{2}}

Solution steps

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

Apply the Product Rule:

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

f = x^{2}, g = sec (\frac{1}{x ^{3}})

= \frac{d}{d x} (x^{2}) sec (\frac{1}{x ^{3}}) + \frac{d}{d x} (sec (\frac{1}{x ^{3}})) x^{2}

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

\frac{d}{d x} (sec (\frac{1}{x ^{3}})) = - \frac{3 sec ( \frac{1}{x ^{3}} ) tan ( \frac{1}{x ^{3}} )}{x ^{4}}

= 2 x sec (\frac{1}{x ^{3}}) + (- \frac{3 sec ( \frac{1}{x ^{3}} ) tan ( \frac{1}{x ^{3}} )}{x ^{4}}) x^{2}

Simplify

2 x sec (\frac{1}{x ^{3}}) + (- \frac{3 sec ( \frac{1}{x ^{3}} ) tan ( \frac{1}{x ^{3}} )}{x ^{4}}) x^{2} : 2 x sec (\frac{1}{x ^{3}}) - \frac{3 sec ( \frac{1}{x ^{3}} ) tan ( \frac{1}{x ^{3}} )}{x ^{2}}

= 2 x sec (\frac{1}{x ^{3}}) - \frac{3 sec ( \frac{1}{x ^{3}} ) tan ( \frac{1}{x ^{3}} )}{x ^{2}}

\int \frac{x ^{3}}{x ^{8} + 5} d x

\frac{\partial}{\partial x} (e^{2 y} (3 e^{y} - 2 x))

\frac{s + 1}{( s + 6 ) ^{2} \cdot ( s + 2 )}

\frac{d}{d x} (- e^{x - 28} + 75)

\int - x - 2 d x

What is the derivative of x^2sec(1/(x^3)) ?
The derivative of x^2sec(1/(x^3)) is 2xsec(1/(x^3))-(3sec(1/(x^3))tan(1/(x^3)))/(x^2)
What is the first derivative of x^2sec(1/(x^3)) ?
The first derivative of x^2sec(1/(x^3)) is 2xsec(1/(x^3))-(3sec(1/(x^3))tan(1/(x^3)))/(x^2)