What is the derivative of (8t)/(t^2+9) ?

The derivative of (8t)/(t^2+9) is (8(-t^2+9))/((t^2+9)^2)

What is the first derivative of (8t)/(t^2+9) ?

The first derivative of (8t)/(t^2+9) is (8(-t^2+9))/((t^2+9)^2)

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derivative of

\frac{8 t}{t ^{2} + 9}

\frac{8 ( - t ^{2} + 9 )}{( t ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{8 t}{t ^{2} + 9})

Take the constant out:

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

= 8 \frac{d}{d t} (\frac{t}{t ^{2} + 9})

Apply the Quotient Rule:

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

= 8 \cdot \frac{\frac{d t}{d t} ( t ^{2} + 9 ) - \frac{d}{d t} ( t ^{2} + 9 ) t}{( t ^{2} + 9 ) ^{2}}

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

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

= 8 \cdot \frac{1 \cdot ( t ^{2} + 9 ) - 2 tt}{( t ^{2} + 9 ) ^{2}}

Simplify

8 \cdot \frac{1 \cdot ( t ^{2} + 9 ) - 2 tt}{( t ^{2} + 9 ) ^{2}} : \frac{8 ( - t ^{2} + 9 )}{( t ^{2} + 9 ) ^{2}}

= \frac{8 ( - t ^{2} + 9 )}{( t ^{2} + 9 ) ^{2}}

\frac{( x - 1 )}{( x + 1 )}

\frac{d}{d x} (e^{0.5 x^{2}})

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

n = 2 \sum \infty 5 (- \frac{3}{4})^{n}

f (x) = e^{- \frac{3}{t ^{2}}}

What is the derivative of (8t)/(t^2+9) ?
The derivative of (8t)/(t^2+9) is (8(-t^2+9))/((t^2+9)^2)
What is the first derivative of (8t)/(t^2+9) ?
The first derivative of (8t)/(t^2+9) is (8(-t^2+9))/((t^2+9)^2)