What is the derivative of (tsin(t))/(1+t) ?

The derivative of (tsin(t))/(1+t) is (t^2cos(t)+tcos(t)+sin(t))/((1+t)^2)

What is the first derivative of (tsin(t))/(1+t) ?

The first derivative of (tsin(t))/(1+t) is (t^2cos(t)+tcos(t)+sin(t))/((1+t)^2)

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derivative of

\frac{t sin ( t )}{1 + t}

\frac{t ^{2} cos ( t ) + t cos ( t ) + sin ( t )}{( 1 + t ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{t sin ( t )}{1 + t})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( t sin ( t ) ) ( 1 + t ) - \frac{d}{d t} ( 1 + t ) t sin ( t )}{( 1 + t ) ^{2}}

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

\frac{d}{d t} (1 + t) = 1

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

Simplify

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

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

\frac{x}{1 + x ^{2}}, \frac{( x ^{2} )}{1 + x ^{3}}

y^{^{'}} = \frac{x y + 2 y}{x + 3}

x (x^{3} + 1) y^{^{'}} + (2 x^{3} - 1) y = \frac{x ^{3} - 2}{x}

\int x^{2} - 49 d x

y^{^{''}} + 2 y^{^{'}} + 2 = 0

What is the derivative of (tsin(t))/(1+t) ?
The derivative of (tsin(t))/(1+t) is (t^2cos(t)+tcos(t)+sin(t))/((1+t)^2)
What is the first derivative of (tsin(t))/(1+t) ?
The first derivative of (tsin(t))/(1+t) is (t^2cos(t)+tcos(t)+sin(t))/((1+t)^2)