What is the derivative of (1+tan^2(x)/(xsec(x))) ?

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x sec (x)) = sec (x) + x sec (x) tan (x)

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

Simplify

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

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

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