What is (\partial)/(\partial t)(2e^{sin(x+2ct)}) ?

The answer to (\partial)/(\partial t)(2e^{sin(x+2ct)}) is 4ce^{sin(x+2ct)}cos(x+2ct)

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(\partial)/(\partial t)(2e^{sin(x+2ct)})

\frac{\partial}{\partial t} (2 e^{s i n (x + 2 c t)})

4 c e^{s i n (x + 2 c t)} cos (x + 2 c t)

Solution steps

\frac{\partial}{\partial t} (2 e^{s i n (x + 2 c t)})

Treat

c, x

as a constant

Take the constant out:

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

= 2 \frac{\partial}{\partial t} (e^{s i n (x + 2 c t)})

Apply the chain rule:

e^{s i n (x + 2 c t)} \frac{\partial}{\partial t} (sin (x + 2 c t))

= e^{s i n (x + 2 c t)} \frac{\partial}{\partial t} (sin (x + 2 c t))

\frac{\partial}{\partial t} (sin (x + 2 c t)) = cos (x + 2 c t) \cdot 2 c

= 2 e^{s i n (x + 2 c t)} cos (x + 2 c t) \cdot 2 c

Simplify

= 4 c e^{s i n (x + 2 c t)} cos (x + 2 c t)

\frac{d x}{d t} = - 2 x

\int cos^{2} (2 t) d t

y^{^{''}} + 9 = 0

- 2 π sin (4 π x)

\frac{d}{d x} (230 - 8 x + 0.04 x^{2})

What is (\partial)/(\partial t)(2e^{sin(x+2ct)}) ?
The answer to (\partial)/(\partial t)(2e^{sin(x+2ct)}) is 4ce^{sin(x+2ct)}cos(x+2ct)