What is the derivative of f(t)=((t^2-4t))/(t+3) ?

The derivative of f(t)=((t^2-4t))/(t+3) is (t^2+6t-12)/((t+3)^2)

What is the first derivative of f(t)=((t^2-4t))/(t+3) ?

The first derivative of f(t)=((t^2-4t))/(t+3) is (t^2+6t-12)/((t+3)^2)

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derivative of

f (t) = \frac{( t ^{2} - 4 t )}{t + 3}

\frac{t ^{2} + 6 t - 12}{( t + 3 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{t ^{2} - 4 t}{t + 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( t ^{2} - 4 t ) ( t + 3 ) - \frac{d}{d t} ( t + 3 ) ( t ^{2} - 4 t )}{( t + 3 ) ^{2}}

\frac{d}{d t} (t^{2} - 4 t) = 2 t - 4

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

= \frac{( 2 t - 4 ) ( t + 3 ) - 1 \cdot ( t ^{2} - 4 t )}{( t + 3 ) ^{2}}

Simplify

\frac{( 2 t - 4 ) ( t + 3 ) - 1 \cdot ( t ^{2} - 4 t )}{( t + 3 ) ^{2}} : \frac{t ^{2} + 6 t - 12}{( t + 3 ) ^{2}}

= \frac{t ^{2} + 6 t - 12}{( t + 3 ) ^{2}}

\int \frac{2 x + 3}{x ^{2} + 3 x + 4} d x

x \to 0 + lim (x^{n})

\frac{\partial}{\partial x} (8 x e^{x^{2}} + y^{2})

\int \frac{x + 4}{x ^{2} + 4} d x

What is the derivative of f(t)=((t^2-4t))/(t+3) ?
The derivative of f(t)=((t^2-4t))/(t+3) is (t^2+6t-12)/((t+3)^2)
What is the first derivative of f(t)=((t^2-4t))/(t+3) ?
The first derivative of f(t)=((t^2-4t))/(t+3) is (t^2+6t-12)/((t+3)^2)