What is the derivative of (e^{1-cos(x})/(cos(x))) ?

The derivative of (e^{1-cos(x})/(cos(x))) is e^{1-cos(x)}tan(x)+e^{1-cos(x)}sec(x)tan(x)

What is the first derivative of (e^{1-cos(x})/(cos(x))) ?

The first derivative of (e^{1-cos(x})/(cos(x))) is e^{1-cos(x)}tan(x)+e^{1-cos(x)}sec(x)tan(x)

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\frac{d}{d x} (\frac{e ^{1 - c o s (x)}}{cos ( x )})

e^{1 - c o s (x)} tan (x) + e^{1 - c o s (x)} sec (x) tan (x)

Solution steps

\frac{d}{d x} (\frac{e ^{1 - c o s (x)}}{cos ( x )})

Simplify

\frac{e ^{1 - c o s (x)}}{cos ( x )} : e^{1 - c o s (x)} sec (x)

= \frac{d}{d x} (e^{1 - c o s (x)} sec (x))

Apply the Product Rule:

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

f = e^{1 - c o s (x)}, g = sec (x)

= \frac{d}{d x} (e^{1 - c o s (x)}) sec (x) + \frac{d}{d x} (sec (x)) e^{1 - c o s (x)}

\frac{d}{d x} (e^{1 - c o s (x)}) = e^{1 - c o s (x)} sin (x)

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

= e^{1 - c o s (x)} sin (x) sec (x) + sec (x) tan (x) e^{1 - c o s (x)}

sec (x) sin (x) = tan (x)

= e^{1 - c o s (x)} tan (x) + e^{1 - c o s (x)} sec (x) tan (x)

\frac{d}{d u} (cos (c) (cot (u)))

y^{^{'}} = e^{2 x + y}

\int \frac{5}{5 + e ^{x}} d x

\frac{\partial}{\partial x} (ln (1 - e^{- x}))

\frac{d}{d x} (arctan (\frac{x}{a}))

What is the derivative of (e^{1-cos(x})/(cos(x))) ?
The derivative of (e^{1-cos(x})/(cos(x))) is e^{1-cos(x)}tan(x)+e^{1-cos(x)}sec(x)tan(x)
What is the first derivative of (e^{1-cos(x})/(cos(x))) ?
The first derivative of (e^{1-cos(x})/(cos(x))) is e^{1-cos(x)}tan(x)+e^{1-cos(x)}sec(x)tan(x)