What is (\partial)/(\partial t)(9cos(t)) ?

The answer to (\partial)/(\partial t)(9cos(t)) is -9sin(t)

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(\partial)/(\partial t)(9cos(t))

\frac{\partial}{\partial t} (9 cos (t))

- 9 sin (t)

Solution steps

\frac{\partial}{\partial t} (9 cos (t))

Take the constant out:

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

= 9 \frac{\partial}{\partial t} (cos (t))

Apply the common derivative:

\frac{\partial}{\partial t} (cos (t)) = - sin (t)

= 9 (- sin (t))

Simplify

= - 9 sin (t)

\frac{d}{d x} (\frac{x ^{3} ln ( x )}{x + 2})

\int \frac{3}{x x ^{2} - 1} d x

\frac{d}{d x} f (x) = 4 x - 9 x^{2}

\int_{2}^{4} x (\frac{( 2 x - 3 )}{6}) d x

\int \frac{5 x + 14}{( x + 1 ) ( x ^{2} - 4 )} d x

What is (\partial)/(\partial t)(9cos(t)) ?
The answer to (\partial)/(\partial t)(9cos(t)) is -9sin(t)