What is the derivative of e^{cos(2x}sin(2x)) ?

The derivative of e^{cos(2x}sin(2x)) is -2e^{cos(2x)}sin^2(2x)+2e^{cos(2x)}cos(2x)

What is the first derivative of e^{cos(2x}sin(2x)) ?

The first derivative of e^{cos(2x}sin(2x)) is -2e^{cos(2x)}sin^2(2x)+2e^{cos(2x)}cos(2x)

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\frac{d}{d x} (e^{c o s (2 x)} sin (2 x))

- 2 e^{c o s (2 x)} sin^{2} (2 x) + 2 e^{c o s (2 x)} cos (2 x)

Solution steps

\frac{d}{d x} (e^{c o s (2 x)} sin (2 x))

Apply the Product Rule:

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

f = e^{c o s (2 x)}, g = sin (2 x)

= \frac{d}{d x} (e^{c o s (2 x)}) sin (2 x) + \frac{d}{d x} (sin (2 x)) e^{c o s (2 x)}

\frac{d}{d x} (e^{c o s (2 x)}) = - 2 e^{c o s (2 x)} sin (2 x)

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

= (- 2 e^{c o s (2 x)} sin (2 x)) sin (2 x) + cos (2 x) \cdot 2 e^{c o s (2 x)}

Simplify

(- 2 e^{c o s (2 x)} sin (2 x)) sin (2 x) + cos (2 x) \cdot 2 e^{c o s (2 x)} : - 2 e^{c o s (2 x)} sin^{2} (2 x) + 2 e^{c o s (2 x)} cos (2 x)

= - 2 e^{c o s (2 x)} sin^{2} (2 x) + 2 e^{c o s (2 x)} cos (2 x)

x \to \infty lim (\frac{2 e ^{2 x}}{2 x})

x \to 0 lim (cos (\frac{9}{x}))

\frac{d}{d x} (cos (x - 2))

\frac{d}{d t} (\frac{t}{3})

\int \frac{16 - 16 x}{1 - x} d x

What is the derivative of e^{cos(2x}sin(2x)) ?
The derivative of e^{cos(2x}sin(2x)) is -2e^{cos(2x)}sin^2(2x)+2e^{cos(2x)}cos(2x)
What is the first derivative of e^{cos(2x}sin(2x)) ?
The first derivative of e^{cos(2x}sin(2x)) is -2e^{cos(2x)}sin^2(2x)+2e^{cos(2x)}cos(2x)