What is the derivative of (5t)/(1+2t^2) ?

The derivative of (5t)/(1+2t^2) is (5(1-2t^2))/((1+2t^2)^2)

What is the first derivative of (5t)/(1+2t^2) ?

The first derivative of (5t)/(1+2t^2) is (5(1-2t^2))/((1+2t^2)^2)

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derivative of

\frac{5 t}{1 + 2 t ^{2}}

\frac{5 ( 1 - 2 t ^{2} )}{( 1 + 2 t ^{2} ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{5 t}{1 + 2 t ^{2}})

Take the constant out:

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

= 5 \frac{d}{d t} (\frac{t}{1 + 2 t ^{2}})

Apply the Quotient Rule:

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

= 5 \cdot \frac{\frac{d t}{d t} ( 1 + 2 t ^{2} ) - \frac{d}{d t} ( 1 + 2 t ^{2} ) t}{( 1 + 2 t ^{2} ) ^{2}}

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

\frac{d}{d t} (1 + 2 t^{2}) = 4 t

= 5 \cdot \frac{1 \cdot ( 1 + 2 t ^{2} ) - 4 tt}{( 1 + 2 t ^{2} ) ^{2}}

Simplify

5 \cdot \frac{1 \cdot ( 1 + 2 t ^{2} ) - 4 tt}{( 1 + 2 t ^{2} ) ^{2}} : \frac{5 ( 1 - 2 t ^{2} )}{( 1 + 2 t ^{2} ) ^{2}}

= \frac{5 ( 1 - 2 t ^{2} )}{( 1 + 2 t ^{2} ) ^{2}}

f (x) = 7 x + 1, (9, 8)

\int \frac{a}{a ^{2} + x ^{2}} d x

\int \frac{1}{5 x + 4} d x

n = 1 \sum \infty \frac{8 n}{7 - 4 n}

\frac{d}{d x} (3 tan (2^{- x}))

What is the derivative of (5t)/(1+2t^2) ?
The derivative of (5t)/(1+2t^2) is (5(1-2t^2))/((1+2t^2)^2)
What is the first derivative of (5t)/(1+2t^2) ?
The first derivative of (5t)/(1+2t^2) is (5(1-2t^2))/((1+2t^2)^2)