What is the derivative of 2sin(4pi(x+1/8)) ?

The derivative of 2sin(4pi(x+1/8)) is 8picos(4pi(x+1/8))

What is the first derivative of 2sin(4pi(x+1/8)) ?

The first derivative of 2sin(4pi(x+1/8)) is 8picos(4pi(x+1/8))

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derivative of

2 sin (4 π (x + \frac{1}{8}))

8 π cos (4 π (x + \frac{1}{8}))

Solution steps

\frac{d}{d x} (2 sin (4 π (x + \frac{1}{8})))

Take the constant out:

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

= 2 \frac{d}{d x} (sin (4 π (x + \frac{1}{8})))

Apply the chain rule:

cos (4 π (x + \frac{1}{8})) \frac{d}{d x} (4 π (x + \frac{1}{8}))

= cos (4 π (x + \frac{1}{8})) \frac{d}{d x} (4 π (x + \frac{1}{8}))

\frac{d}{d x} (4 π (x + \frac{1}{8})) = 4 π

= 2 cos (4 π (x + \frac{1}{8})) \cdot 4 π

Simplify

= 8 π cos (4 π (x + \frac{1}{8}))

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

x \to - 1 lim (8 x^{3})

4 x^{2}, x^{4} - 6 x^{2}, [- 3, 3]

x \to 0 lim (\frac{sin ( π + x )}{x})

y = 12, y = 2 x^{2} - 11 x + 12

What is the derivative of 2sin(4pi(x+1/8)) ?
The derivative of 2sin(4pi(x+1/8)) is 8picos(4pi(x+1/8))
What is the first derivative of 2sin(4pi(x+1/8)) ?
The first derivative of 2sin(4pi(x+1/8)) is 8picos(4pi(x+1/8))