What is the (sec(x^5+x))^' ?

The (sec(x^5+x))^' is sec(x^5+x)tan(x^5+x)(5x^4+1)

What is the first (sec(x^5+x))^' ?

The first (sec(x^5+x))^' is sec(x^5+x)tan(x^5+x)(5x^4+1)

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(sec (x^{5} + x))^{^{'}}

sec (x^{5} + x) tan (x^{5} + x) (5 x^{4} + 1)

Solution steps

(sec (x^{5} + x))^{'}

Apply the chain rule:

sec (x^{5} + x) tan (x^{5} + x) (x^{5} + x)^{'}

= sec (x^{5} + x) tan (x^{5} + x) (x^{5} + x)^{'}

(x^{5} + x)^{'} = 5 x^{4} + 1

= sec (x^{5} + x) tan (x^{5} + x) (5 x^{4} + 1)

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

\int \frac{( 4 x + 13 )}{x ^{2} + 3 x - 10} d x

\int_{0}^{\frac{π}{2}} \frac{5 cos ( x )}{sin ( x )} d x

\frac{d y}{d t} = \frac{4 t ^{8}}{e + t ^{9}}

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

What is the (sec(x^5+x))^' ?
The (sec(x^5+x))^' is sec(x^5+x)tan(x^5+x)(5x^4+1)
What is the first (sec(x^5+x))^' ?
The first (sec(x^5+x))^' is sec(x^5+x)tan(x^5+x)(5x^4+1)