What is the derivative of g(t)=6t^3cos(t) ?

The derivative of g(t)=6t^3cos(t) is 6(3t^2cos(t)-t^3sin(t))

What is the first derivative of g(t)=6t^3cos(t) ?

The first derivative of g(t)=6t^3cos(t) is 6(3t^2cos(t)-t^3sin(t))

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derivative of

g (t) = 6 t^{3} cos (t)

6 (3 t^{2} cos (t) - t^{3} sin (t))

Solution steps

\frac{d}{d t} (6 t^{3} cos (t))

Take the constant out:

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

= 6 \frac{d}{d t} (t^{3} cos (t))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{3}, g = cos (t)

= 6 (\frac{d}{d t} (t^{3}) cos (t) + \frac{d}{d t} (cos (t)) t^{3})

\frac{d}{d t} (t^{3}) = 3 t^{2}

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

= 6 (3 t^{2} cos (t) + (- sin (t)) t^{3})

Simplify

= 6 (3 t^{2} cos (t) - t^{3} sin (t))

f (x) = x y

\int_{1}^{4} x^{\frac{3}{2}} d x

\int sin (3) (x) d x

\frac{d}{d x} ((x^{2} + y^{2})^{\frac{1}{2}})

x^{6} + y^{5} x^{7} + 4 y^{^{'}} = 0

What is the derivative of g(t)=6t^3cos(t) ?
The derivative of g(t)=6t^3cos(t) is 6(3t^2cos(t)-t^3sin(t))
What is the first derivative of g(t)=6t^3cos(t) ?
The first derivative of g(t)=6t^3cos(t) is 6(3t^2cos(t)-t^3sin(t))