What is (\partial)/(\partial x)(acos^2(2pi)(t-x)) ?

The answer to (\partial)/(\partial x)(acos^2(2pi)(t-x)) is -a

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(\partial)/(\partial x)(acos^2(2pi)(t-x))

\frac{\partial}{\partial x} (a cos^{2} (2 π) (t - x))

- a

Solution steps

\frac{\partial}{\partial x} (a cos^{2} (2 π) (t - x))

Simplify

a cos^{2} (2 π) (t - x) : a (t - x)

= \frac{\partial}{\partial x} (a (t - x))

Treat

a, t

as a constant

Take the constant out:

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

= a \frac{\partial}{\partial x} (t - x)

Apply the Sum/Difference Rule:

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

= a (\frac{\partial}{\partial x} (t) - \frac{\partial}{\partial x} (x))

\frac{\partial}{\partial x} (t) = 0

\frac{\partial}{\partial x} (x) = 1

= a (0 - 1)

Simplify

= - a

\frac{x ^{2}}{x - 5}

f (x) = x^{2} + 7

\frac{d}{d x} (x^{0.1})

\frac{\partial}{\partial y} (arctan (y))

y = x^{2} + 3, (- 2, 4)

What is (\partial)/(\partial x)(acos^2(2pi)(t-x)) ?
The answer to (\partial)/(\partial x)(acos^2(2pi)(t-x)) is -a