What is (\partial}{\partial t}(\frac{sin(t))/t) ?

The answer to (\partial}{\partial t}(\frac{sin(t))/t) is (tcos(t)-sin(t))/(t^2)

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(\partial}{\partial t}(\frac{sin(t))/t)

\frac{\partial}{\partial t} (\frac{sin ( t )}{t})

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

Solution steps

\frac{\partial}{\partial t} (\frac{sin ( t )}{t})

Apply the Quotient Rule:

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

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

\frac{\partial}{\partial t} (sin (t)) = cos (t)

\frac{\partial}{\partial t} (t) = 1

= \frac{cos ( t ) t - 1 \cdot sin ( t )}{t ^{2}}

Simplify

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

n = 1 \sum \infty (0.1)^{n - 1}

\frac{\partial}{\partial x} (tan (4 x))

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

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

x y^{^{'}} + 3 y = 5 x^{2}

What is (\partial}{\partial t}(\frac{sin(t))/t) ?
The answer to (\partial}{\partial t}(\frac{sin(t))/t) is (tcos(t)-sin(t))/(t^2)