What is the derivative of y=t^2+5cos(t) ?

The derivative of y=t^2+5cos(t) is 2t-5sin(t)

What is the first derivative of y=t^2+5cos(t) ?

The first derivative of y=t^2+5cos(t) is 2t-5sin(t)

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derivative of

y = t^{2} + 5 cos (t)

2 t - 5 sin (t)

Solution steps

\frac{d}{d t} (t^{2} + 5 cos (t))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{2}) + \frac{d}{d t} (5 cos (t))

\frac{d}{d t} (t^{2}) = 2 t

\frac{d}{d t} (5 cos (t)) = - 5 sin (t)

= 2 t - 5 sin (t)

\frac{d y}{d x} sin (x + y) = 0

\int \frac{sec ( x ) + cos ( x )}{4 cos ( x )} d x

x \to 0 lim (x \cdot lo g_{2} (x))

\int \frac{x ^{2} + 36}{x} d x

t \to 3 lim (4 t^{2} - 2 t + 1)

What is the derivative of y=t^2+5cos(t) ?
The derivative of y=t^2+5cos(t) is 2t-5sin(t)
What is the first derivative of y=t^2+5cos(t) ?
The first derivative of y=t^2+5cos(t) is 2t-5sin(t)