What is the derivative of (2t+1)/(t+3) ?

The derivative of (2t+1)/(t+3) is 5/((t+3)^2)

What is the first derivative of (2t+1)/(t+3) ?

The first derivative of (2t+1)/(t+3) is 5/((t+3)^2)

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derivative of

\frac{2 t + 1}{t + 3}

\frac{5}{( t + 3 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{2 t + 1}{t + 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 2 t + 1 ) ( t + 3 ) - \frac{d}{d t} ( t + 3 ) ( 2 t + 1 )}{( t + 3 ) ^{2}}

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

\frac{d}{d t} (t + 3) = 1

= \frac{2 ( t + 3 ) - 1 \cdot ( 2 t + 1 )}{( t + 3 ) ^{2}}

Simplify

\frac{2 ( t + 3 ) - 1 \cdot ( 2 t + 1 )}{( t + 3 ) ^{2}} : \frac{5}{( t + 3 ) ^{2}}

= \frac{5}{( t + 3 ) ^{2}}

\frac{\partial}{\partial x} (e^{x^{2} - y})

8 x + 9 x e^{x}

\int 2 ln (t) d t

y^{^{'}} = 1 - 7 x^{2}

\int \frac{x - 2}{x + 3} d x

What is the derivative of (2t+1)/(t+3) ?
The derivative of (2t+1)/(t+3) is 5/((t+3)^2)
What is the first derivative of (2t+1)/(t+3) ?
The first derivative of (2t+1)/(t+3) is 5/((t+3)^2)