What is the derivative of 1+sec(x) ?

The derivative of 1+sec(x) is sec(x)tan(x)

What is the first derivative of 1+sec(x) ?

The first derivative of 1+sec(x) is sec(x)tan(x)

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\frac{d}{d x} (1 + sec (x))

sec (x) tan (x)

Solution steps

\frac{d}{d x} (1 + sec (x))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (1) = 0

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

= 0 + sec (x) tan (x)

Simplify

= sec (x) tan (x)

\frac{d}{d x} (- x^{5} - 6 x^{4} - 14 x^{3} - 16 x^{2} - 9 x - 2)

\frac{\partial ^{2}}{\partial x \partial y} (ln (3 + x^{2} y^{2}))

x \to 0 - lim (\frac{sin ( ∣ x ∣ )}{x})

\int (3 + 2 x) d x

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

What is the derivative of 1+sec(x) ?
The derivative of 1+sec(x) is sec(x)tan(x)
What is the first derivative of 1+sec(x) ?
The first derivative of 1+sec(x) is sec(x)tan(x)