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

The answer to derivative of f(x)=(5t+300)/(t^2+25) is (dt)/(dx)=0

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derivative of f(x)=(5t+300)/(t^2+25)

derivative of

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

\frac{d t}{d x} = 0

Solution steps

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

Treat

t

t (x)

Differentiate both sides:

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

Isolate

\frac{d t}{d x} : \frac{d t}{d x} = 0

\frac{d t}{d x} = 0

2 x^{3} + 4 x

\int x^{2} y^{2} d x

(- 7 x^{2} + 7)^{5} (- 7 x^{2} + 4)^{- 4}

ln (1 + 4 x)

What is derivative of f(x)=(5t+300)/(t^2+25) ?
The answer to derivative of f(x)=(5t+300)/(t^2+25) is (dt)/(dx)=0