What is the derivative of f(t)=(cos(t))/(t^3) ?

The derivative of f(t)=(cos(t))/(t^3) is -(tsin(t)+3cos(t))/(t^4)

What is the first derivative of f(t)=(cos(t))/(t^3) ?

The first derivative of f(t)=(cos(t))/(t^3) is -(tsin(t)+3cos(t))/(t^4)

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derivative of

f (t) = \frac{cos ( t )}{t ^{3}}

- \frac{t sin ( t ) + 3 cos ( t )}{t ^{4}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

\frac{( - sin ( t ) ) t ^{3} - 3 t ^{2} cos ( t )}{( t ^{3} ) ^{2}} : - \frac{t sin ( t ) + 3 cos ( t )}{t ^{4}}

= - \frac{t sin ( t ) + 3 cos ( t )}{t ^{4}}

f (t) = sec (t)

n = 0 \sum \infty (\frac{e}{6 π})^{n}

\frac{x ^{2}}{1 - x}

y^{^{''}} + 3 y^{^{'}} - 10 y = 6 e^{4 x}

\int_{0}^{2} \frac{1}{16 + x ^{2}} d x

What is the derivative of f(t)=(cos(t))/(t^3) ?
The derivative of f(t)=(cos(t))/(t^3) is -(tsin(t)+3cos(t))/(t^4)
What is the first derivative of f(t)=(cos(t))/(t^3) ?
The first derivative of f(t)=(cos(t))/(t^3) is -(tsin(t)+3cos(t))/(t^4)