What is the derivative of 4cos(t)-9sin(t) ?

The derivative of 4cos(t)-9sin(t) is -4sin(t)-9cos(t)

What is the first derivative of 4cos(t)-9sin(t) ?

The first derivative of 4cos(t)-9sin(t) is -4sin(t)-9cos(t)

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derivative of

4 cos (t) - 9 sin (t)

- 4 sin (t) - 9 cos (t)

Solution steps

\frac{d}{d t} (4 cos (t) - 9 sin (t))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (4 cos (t)) - \frac{d}{d t} (9 sin (t))

\frac{d}{d t} (4 cos (t)) = - 4 sin (t)

\frac{d}{d t} (9 sin (t)) = 9 cos (t)

= - 4 sin (t) - 9 cos (t)

\frac{\partial}{\partial x} (x^{9} y z^{2} + 5 yz)

\int cos^{3} (2 x) sin^{3} (2 x) d x

cos (y) y^{^{'}} = e^{2 x} + 1

\frac{y ^{3}}{30} + \frac{5}{2 y}

ln (1 + x)

What is the derivative of 4cos(t)-9sin(t) ?
The derivative of 4cos(t)-9sin(t) is -4sin(t)-9cos(t)
What is the first derivative of 4cos(t)-9sin(t) ?
The first derivative of 4cos(t)-9sin(t) is -4sin(t)-9cos(t)