What is the derivative of 2sin(2x-2cos(2x)-1/5 e^x) ?

The derivative of 2sin(2x-2cos(2x)-1/5 e^x) is 4cos(2x)+4sin(2x)-1/5 e^x

What is the first derivative of 2sin(2x-2cos(2x)-1/5 e^x) ?

The first derivative of 2sin(2x-2cos(2x)-1/5 e^x) is 4cos(2x)+4sin(2x)-1/5 e^x

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\frac{d}{d x} (2 sin (2 x) - 2 cos (2 x) - \frac{1}{5} e^{x})

4 cos (2 x) + 4 sin (2 x) - \frac{1}{5} e^{x}

Solution steps

\frac{d}{d x} (2 sin (2 x) - 2 cos (2 x) - \frac{1}{5} e^{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 sin (2 x)) - \frac{d}{d x} (2 cos (2 x)) - \frac{d}{d x} (\frac{1}{5} e^{x})

\frac{d}{d x} (2 sin (2 x)) = 4 cos (2 x)

\frac{d}{d x} (2 cos (2 x)) = - 4 sin (2 x)

\frac{d}{d x} (\frac{1}{5} e^{x}) = \frac{1}{5} e^{x}

= 4 cos (2 x) - (- 4 sin (2 x)) - \frac{1}{5} e^{x}

Simplify

= 4 cos (2 x) + 4 sin (2 x) - \frac{1}{5} e^{x}

\frac{d}{d x} ((x - 5)^{2})

n = 0 \sum \infty (- 1)^{n} \cdot n

y = - 80 + 208

\frac{\partial}{\partial y} (tan (x) y^{2.5})

\frac{d}{d x} ((x^{3} + 2 x) e^{s i n (x)})

What is the derivative of 2sin(2x-2cos(2x)-1/5 e^x) ?
The derivative of 2sin(2x-2cos(2x)-1/5 e^x) is 4cos(2x)+4sin(2x)-1/5 e^x
What is the first derivative of 2sin(2x-2cos(2x)-1/5 e^x) ?
The first derivative of 2sin(2x-2cos(2x)-1/5 e^x) is 4cos(2x)+4sin(2x)-1/5 e^x