What is the derivative of g(x)=tan(2x)*sin(2x) ?

The derivative of g(x)=tan(2x)*sin(2x) is 2sin(2x)+2sec(2x)tan(2x)

What is the first derivative of g(x)=tan(2x)*sin(2x) ?

The first derivative of g(x)=tan(2x)*sin(2x) is 2sin(2x)+2sec(2x)tan(2x)

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derivative of

g (x) = tan (2 x) \cdot sin (2 x)

2 sin (2 x) + 2 sec (2 x) tan (2 x)

Solution steps

\frac{d}{d x} (tan (2 x) sin (2 x))

Apply the Product Rule:

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

f = tan (2 x), g = sin (2 x)

= \frac{d}{d x} (tan (2 x)) sin (2 x) + \frac{d}{d x} (sin (2 x)) tan (2 x)

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

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

= sec^{2} (2 x) \cdot 2 sin (2 x) + cos (2 x) \cdot 2 tan (2 x)

Simplify

sec^{2} (2 x) \cdot 2 sin (2 x) + cos (2 x) \cdot 2 tan (2 x) : 2 sin (2 x) + 2 sec (2 x) tan (2 x)

= 2 sin (2 x) + 2 sec (2 x) tan (2 x)

(2 y^{2} x - y) d x + x d y = 0

y = 2 x - x^{2}

\frac{\partial}{\partial x} (\frac{10000 x}{y ^{2}})

n = 1 \sum \infty (\frac{8}{5})^{n - 1}

(t^{2} + 1) f^{^{'}} (t) + [f (t)]^{2} + 1 = 0, f (4) = e

What is the derivative of g(x)=tan(2x)*sin(2x) ?
The derivative of g(x)=tan(2x)*sin(2x) is 2sin(2x)+2sec(2x)tan(2x)
What is the first derivative of g(x)=tan(2x)*sin(2x) ?
The first derivative of g(x)=tan(2x)*sin(2x) is 2sin(2x)+2sec(2x)tan(2x)