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Popular >

d/(dt)(t/(1+t^2))

Solution

\frac{d}{d t} (\frac{t}{1 + t ^{2}})

Solution

\frac{1 - t ^{2}}{( 1 + t ^{2} ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{t}{1 + t ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d t}{d t} ( 1 + t ^{2} ) - \frac{d}{d t} ( 1 + t ^{2} ) t}{( 1 + t ^{2} ) ^{2}}

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

\frac{d}{d t} (1 + t^{2}) = 2 t

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

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

= \frac{1 - t ^{2}}{( 1 + t ^{2} ) ^{2}}

Graph

Popular Examples

area x^2,1,0 area

x^{2}, 1, 0

tangent of f(x)=9x^2+7x-6 tangent of

f (x) = 9 x^{2} + 7 x - 6

(\partial)/(\partial x)(-3xy^2-2ycos(x))

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

integral of 1/(t^2sqrt(169-t^2))

\int \frac{1}{t ^{2} 169 - t ^{2}} d t

normal of y=(sqrt(x))/(x+6),(4,0.2)normal of

y = \frac{x}{x + 6}, (4, 0.2)

Frequently Asked Questions (FAQ)

What is the d/(dt)(t/(1+t^2)) ?
The d/(dt)(t/(1+t^2)) is (1-t^2)/((1+t^2)^2)
What is the first d/(dt)(t/(1+t^2)) ?
The first d/(dt)(t/(1+t^2)) is (1-t^2)/((1+t^2)^2)