What is the derivative of (4t)/(t^2+25) ?

The derivative of (4t)/(t^2+25) is (4(-t^2+25))/((t^2+25)^2)

What is the first derivative of (4t)/(t^2+25) ?

The first derivative of (4t)/(t^2+25) is (4(-t^2+25))/((t^2+25)^2)

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derivative of

\frac{4 t}{t ^{2} + 25}

\frac{4 ( - t ^{2} + 25 )}{( t ^{2} + 25 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{4 t}{t ^{2} + 25})

Take the constant out:

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

= 4 \frac{d}{d t} (\frac{t}{t ^{2} + 25})

Apply the Quotient Rule:

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

= 4 \cdot \frac{\frac{d t}{d t} ( t ^{2} + 25 ) - \frac{d}{d t} ( t ^{2} + 25 ) t}{( t ^{2} + 25 ) ^{2}}

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

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

= 4 \cdot \frac{1 \cdot ( t ^{2} + 25 ) - 2 tt}{( t ^{2} + 25 ) ^{2}}

Simplify

4 \cdot \frac{1 \cdot ( t ^{2} + 25 ) - 2 tt}{( t ^{2} + 25 ) ^{2}} : \frac{4 ( - t ^{2} + 25 )}{( t ^{2} + 25 ) ^{2}}

= \frac{4 ( - t ^{2} + 25 )}{( t ^{2} + 25 ) ^{2}}

2 sin (2 π x)

sec (\frac{1}{x})

\frac{\partial}{\partial x} (ln (x + 4 y))

\int sin (7 x) sin (9 x) d x

y^{^{'}} + \frac{1 - 2 x}{x ^{2}} y = 1

What is the derivative of (4t)/(t^2+25) ?
The derivative of (4t)/(t^2+25) is (4(-t^2+25))/((t^2+25)^2)
What is the first derivative of (4t)/(t^2+25) ?
The first derivative of (4t)/(t^2+25) is (4(-t^2+25))/((t^2+25)^2)