What is the first (5cos(t))^' ?

The first (5cos(t))^' is -5sin(t)

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(5cos(t))^'

(5 cos (t))^{^{'}}

- 5 sin (t)

Solution steps

(5 cos (t))^{'}

Take the constant out:

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

= 5 (cos (t))^{^{'}}

Apply the common derivative:

(cos (t))^{'} = - sin (t)

= 5 (- sin (t))

Simplify

= - 5 sin (t)

d er i v a t i v e y = \frac{6}{x ^{4}}

x \to - 3 lim (8)

\frac{d}{d x} (cos (h) (x))

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 0