What is the derivative of (9t)/(4+t^2) ?

The derivative of (9t)/(4+t^2) is (9(4-t^2))/((4+t^2)^2)

What is the first derivative of (9t)/(4+t^2) ?

The first derivative of (9t)/(4+t^2) is (9(4-t^2))/((4+t^2)^2)

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derivative of

\frac{9 t}{4 + t ^{2}}

\frac{9 ( 4 - t ^{2} )}{( 4 + t ^{2} ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{9 t}{4 + t ^{2}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 9 \frac{d}{d t} (\frac{t}{4 + t ^{2}})

Apply the Quotient Rule:

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

= 9 \cdot \frac{\frac{d t}{d t} ( 4 + t ^{2} ) - \frac{d}{d t} ( 4 + t ^{2} ) t}{( 4 + t ^{2} ) ^{2}}

\frac{d t}{d t} = 1

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

= 9 \cdot \frac{1 \cdot ( 4 + t ^{2} ) - 2 tt}{( 4 + t ^{2} ) ^{2}}

Simplify

9 \cdot \frac{1 \cdot ( 4 + t ^{2} ) - 2 tt}{( 4 + t ^{2} ) ^{2}} : \frac{9 ( 4 - t ^{2} )}{( 4 + t ^{2} ) ^{2}}

= \frac{9 ( 4 - t ^{2} )}{( 4 + t ^{2} ) ^{2}}

n = 3 \sum \infty \frac{1}{n}

3 - 5 x

\frac{d}{d x} (\frac{3 x}{t - 3})

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

\frac{\partial}{\partial x} (2 \cdot (y + 4) \cdot (x - 4)^{3})

What is the derivative of (9t)/(4+t^2) ?
The derivative of (9t)/(4+t^2) is (9(4-t^2))/((4+t^2)^2)
What is the first derivative of (9t)/(4+t^2) ?
The first derivative of (9t)/(4+t^2) is (9(4-t^2))/((4+t^2)^2)