What is the derivative of (e^x/(2-tan(x))) ?

The derivative of (e^x/(2-tan(x))) is (e^x(2-tan(x))+e^xsec^2(x))/((2-tan(x))^2)

What is the first derivative of (e^x/(2-tan(x))) ?

The first derivative of (e^x/(2-tan(x))) is (e^x(2-tan(x))+e^xsec^2(x))/((2-tan(x))^2)

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\frac{d}{d x} (\frac{e ^{x}}{2 - tan ( x )})

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

Solution steps

\frac{d}{d x} (\frac{e ^{x}}{2 - tan ( x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( e ^{x} ) ( 2 - tan ( x ) ) - \frac{d}{d x} ( 2 - tan ( x ) ) e ^{x}}{( 2 - tan ( x ) ) ^{2}}

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

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

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

Simplify

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

\frac{\partial}{\partial x} (sin (3 x) cos (3 y))

f (x) = 5 x - 3 x^{2}, at x = 1

\frac{d}{d x} (\frac{( x ^{2} - 6 ) ^{3}}{x ^{2} + 4})

\frac{d ^{2}}{d x ^{2}} (x^{\frac{9}{5}} y^{7} z^{\frac{2}{7}})

y^{^{''}} - 6 y^{^{'}} + 9 y = 6 e^{3 x}

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