What is the d/(dt)((3t^2)/(sqrt(1+4t^2+9t^4))) ?

The d/(dt)((3t^2)/(sqrt(1+4t^2+9t^4))) is (3(4t^3+2t))/((1+4t^2+9t^4)sqrt(1+4t^2+9t^4))

What is the first d/(dt)((3t^2)/(sqrt(1+4t^2+9t^4))) ?

The first d/(dt)((3t^2)/(sqrt(1+4t^2+9t^4))) is (3(4t^3+2t))/((1+4t^2+9t^4)sqrt(1+4t^2+9t^4))

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\frac{d}{d t} (\frac{3 t ^{2}}{1 + 4 t ^{2} + 9 t ^{4}})

\frac{3 ( 4 t ^{3} + 2 t )}{( 1 + 4 t ^{2} + 9 t ^{4} ) 1 + 4 t ^{2} + 9 t ^{4}}

Solution steps

\frac{d}{d t} (\frac{3 t ^{2}}{1 + 4 t ^{2} + 9 t ^{4}})

Take the constant out:

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

= 3 \frac{d}{d t} (\frac{t ^{2}}{1 + 4 t ^{2} + 9 t ^{4}})

Apply the Quotient Rule:

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

= 3 \cdot \frac{\frac{d}{d t} ( t ^{2} ) 1 + 4 t ^{2} + 9 t ^{4} - \frac{d}{d t} ( 1 + 4 t ^{2} + 9 t ^{4} ) t ^{2}}{( 1 + 4 t ^{2} + 9 t ^{4} ) ^{2}}

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

\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}}

= 3 \cdot \frac{2 t 1 + 4 t ^{2} + 9 t ^{4} - \frac{2 t ( 9 t ^{2} + 2 )}{1 + 4 t ^{2} + 9 t ^{4}} t ^{2}}{( 1 + 4 t ^{2} + 9 t ^{4} ) ^{2}}

Simplify

3 \cdot \frac{2 t 1 + 4 t ^{2} + 9 t ^{4} - \frac{2 t ( 9 t ^{2} + 2 )}{1 + 4 t ^{2} + 9 t ^{4}} t ^{2}}{( 1 + 4 t ^{2} + 9 t ^{4} ) ^{2}} : \frac{3 ( 4 t ^{3} + 2 t )}{( 1 + 4 t ^{2} + 9 t ^{4} ) 1 + 4 t ^{2} + 9 t ^{4}}

= \frac{3 ( 4 t ^{3} + 2 t )}{( 1 + 4 t ^{2} + 9 t ^{4} ) 1 + 4 t ^{2} + 9 t ^{4}}

\int e^{x^{2}} x^{3} d x

\frac{d}{d x} (a r (x) cos (e^{x}))

\int_{0}^{4} 2 π (2 - y) (4 y^{2}) d y

\int (25 - x^{2} - 3) d x

n = 3 \sum \infty (\frac{3}{11})^{- n}

What is the d/(dt)((3t^2)/(sqrt(1+4t^2+9t^4))) ?
The d/(dt)((3t^2)/(sqrt(1+4t^2+9t^4))) is (3(4t^3+2t))/((1+4t^2+9t^4)sqrt(1+4t^2+9t^4))
What is the first d/(dt)((3t^2)/(sqrt(1+4t^2+9t^4))) ?
The first d/(dt)((3t^2)/(sqrt(1+4t^2+9t^4))) is (3(4t^3+2t))/((1+4t^2+9t^4)sqrt(1+4t^2+9t^4))