What is the derivative of y=(0.9e^x)/(tan(x)) ?

The derivative of y=(0.9e^x)/(tan(x)) is 0.9(e^xcot(x)-e^xcsc^2(x))

What is the first derivative of y=(0.9e^x)/(tan(x)) ?

The first derivative of y=(0.9e^x)/(tan(x)) is 0.9(e^xcot(x)-e^xcsc^2(x))

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derivative of

y = \frac{0.9 e ^{x}}{tan ( x )}

0.9 (e^{x} cot (x) - e^{x} csc^{2} (x))

Solution steps

\frac{d}{d x} (\frac{0.9 e ^{x}}{tan ( x )})

Simplify

\frac{0.9 e ^{x}}{tan ( x )} : 0.9 e^{x} cot (x)

= \frac{d}{d x} (0.9 e^{x} cot (x))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.9 \frac{d}{d x} (e^{x} cot (x))

Apply the Product Rule:

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

f = e^{x}, g = cot (x)

= 0.9 (\frac{d}{d x} (e^{x}) cot (x) + \frac{d}{d x} (cot (x)) e^{x})

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

\frac{d}{d x} (cot (x)) = - csc^{2} (x)

= 0.9 (e^{x} cot (x) + (- csc^{2} (x)) e^{x})

Simplify

= 0.9 (e^{x} cot (x) - e^{x} csc^{2} (x))

f (x) = - 4 sin (2 x) + 3 cos (3 x)

\int \frac{x ^{3} - 4 x ^{2} - 4 x + 20}{x ^{2} - 5} d x

f (x) = 8 e^{- x} + e^{3 x}

\frac{s + 1}{s ^{2}}

f (x) = 1 x^{3}

What is the derivative of y=(0.9e^x)/(tan(x)) ?
The derivative of y=(0.9e^x)/(tan(x)) is 0.9(e^xcot(x)-e^xcsc^2(x))
What is the first derivative of y=(0.9e^x)/(tan(x)) ?
The first derivative of y=(0.9e^x)/(tan(x)) is 0.9(e^xcot(x)-e^xcsc^2(x))