What is the d/(dt)((2t)/(sqrt(1+4t^2+9t^4))) ?

The d/(dt)((2t)/(sqrt(1+4t^2+9t^4))) is (2(-9t^4+1))/((1+4t^2+9t^4)sqrt(1+4t^2+9t^4))

What is the first d/(dt)((2t)/(sqrt(1+4t^2+9t^4))) ?

The first d/(dt)((2t)/(sqrt(1+4t^2+9t^4))) is (2(-9t^4+1))/((1+4t^2+9t^4)sqrt(1+4t^2+9t^4))

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\frac{d}{d t} (\frac{2 t}{1 + 4 t ^{2} + 9 t ^{4}})

\frac{2 ( - 9 t ^{4} + 1 )}{( 1 + 4 t ^{2} + 9 t ^{4} ) 1 + 4 t ^{2} + 9 t ^{4}}

Solution steps

\frac{d}{d t} (\frac{2 t}{1 + 4 t ^{2} + 9 t ^{4}})

Take the constant out:

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

= 2 \frac{d}{d t} (\frac{t}{1 + 4 t ^{2} + 9 t ^{4}})

Apply the Quotient Rule:

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

= 2 \cdot \frac{\frac{d t}{d t} 1 + 4 t ^{2} + 9 t ^{4} - \frac{d}{d t} ( 1 + 4 t ^{2} + 9 t ^{4} ) t}{( 1 + 4 t ^{2} + 9 t ^{4} ) ^{2}}

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

\frac{d}{d t} (1 + 4 t^{2} + 9 t^{4}) = \frac{2 t ( 9 t ^{2} + 2 )}{1 + 4 t ^{2} + 9 t ^{4}}

= 2 \cdot \frac{1 \cdot 1 + 4 t ^{2} + 9 t ^{4} - \frac{2 t ( 9 t ^{2} + 2 )}{1 + 4 t ^{2} + 9 t ^{4}} t}{( 1 + 4 t ^{2} + 9 t ^{4} ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot 1 + 4 t ^{2} + 9 t ^{4} - \frac{2 t ( 9 t ^{2} + 2 )}{1 + 4 t ^{2} + 9 t ^{4}} t}{( 1 + 4 t ^{2} + 9 t ^{4} ) ^{2}} : \frac{2 ( - 9 t ^{4} + 1 )}{( 1 + 4 t ^{2} + 9 t ^{4} ) 1 + 4 t ^{2} + 9 t ^{4}}

= \frac{2 ( - 9 t ^{4} + 1 )}{( 1 + 4 t ^{2} + 9 t ^{4} ) 1 + 4 t ^{2} + 9 t ^{4}}

What is the d/(dt)((2t)/(sqrt(1+4t^2+9t^4))) ?
The d/(dt)((2t)/(sqrt(1+4t^2+9t^4))) is (2(-9t^4+1))/((1+4t^2+9t^4)sqrt(1+4t^2+9t^4))
What is the first d/(dt)((2t)/(sqrt(1+4t^2+9t^4))) ?
The first d/(dt)((2t)/(sqrt(1+4t^2+9t^4))) is (2(-9t^4+1))/((1+4t^2+9t^4)sqrt(1+4t^2+9t^4))