What is the derivative of f(t)=(5t+300)/(t^2+25) ?

The derivative of f(t)=(5t+300)/(t^2+25) is (-5t^2-600t+125)/((t^2+25)^2)

What is the first derivative of f(t)=(5t+300)/(t^2+25) ?

The first derivative of f(t)=(5t+300)/(t^2+25) is (-5t^2-600t+125)/((t^2+25)^2)

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derivative of

f (t) = \frac{5 t + 300}{t ^{2} + 25}

\frac{- 5 t ^{2} - 600 t + 125}{( t ^{2} + 25 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{5 t + 300}{t ^{2} + 25})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 5 t + 300 ) ( t ^{2} + 25 ) - \frac{d}{d t} ( t ^{2} + 25 ) ( 5 t + 300 )}{( t ^{2} + 25 ) ^{2}}

\frac{d}{d t} (5 t + 300) = 5

\frac{d}{d t} (t^{2} + 25) = 2 t

= \frac{5 ( t ^{2} + 25 ) - 2 t ( 5 t + 300 )}{( t ^{2} + 25 ) ^{2}}

Expand

5 (t^{2} + 25) - 2 t (5 t + 300) : - 5 t^{2} - 600 t + 125

= \frac{- 5 t ^{2} - 600 t + 125}{( t ^{2} + 25 ) ^{2}}

\int \frac{2 e ^{2 x}}{e ^{2 x} + 8 e ^{x} + 15} d x

\int \frac{1}{x ^{2} x ^{2} + 5} d x

\int sin (x) + 9 d x

y = x, y = 5 x, y = - x + 2

\int_{0}^{2} x^{3} - x^{2} d x

What is the derivative of f(t)=(5t+300)/(t^2+25) ?
The derivative of f(t)=(5t+300)/(t^2+25) is (-5t^2-600t+125)/((t^2+25)^2)
What is the first derivative of f(t)=(5t+300)/(t^2+25) ?
The first derivative of f(t)=(5t+300)/(t^2+25) is (-5t^2-600t+125)/((t^2+25)^2)