What is the derivative of ((1+16t^2))/((4t)^3) ?

The derivative of ((1+16t^2))/((4t)^3) is (-16t^2-3)/(64t^4)

What is the first derivative of ((1+16t^2))/((4t)^3) ?

The first derivative of ((1+16t^2))/((4t)^3) is (-16t^2-3)/(64t^4)

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derivative of

\frac{( 1 + 16 t ^{2} )}{( 4 t ) ^{3}}

\frac{- 16 t ^{2} - 3}{64 t ^{4}}

Solution steps

\frac{d}{d t} (\frac{1 + 16 t ^{2}}{( 4 t ) ^{3}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 1 + 16 t ^{2} ) ( 4 t ) ^{3} - \frac{d}{d t} ( ( 4 t ) ^{3} ) ( 1 + 16 t ^{2} )}{( ( 4 t ) ^{3} ) ^{2}}

\frac{d}{d t} (1 + 16 t^{2}) = 32 t

\frac{d}{d t} ((4 t)^{3}) = 192 t^{2}

= \frac{32 t ( 4 t ) ^{3} - 192 t ^{2} ( 1 + 16 t ^{2} )}{( ( 4 t ) ^{3} ) ^{2}}

Simplify

\frac{32 t ( 4 t ) ^{3} - 192 t ^{2} ( 1 + 16 t ^{2} )}{( ( 4 t ) ^{3} ) ^{2}} : \frac{- 16 t ^{2} - 3}{64 t ^{4}}

= \frac{- 16 t ^{2} - 3}{64 t ^{4}}

\frac{\partial}{\partial y} (4 x^{3} y^{3} - 3)

\int_{- 1}^{1} x^{2} x^{3} + 1 d x

\int \frac{x ^{2} + 3 x}{x + 4} d x

\int (\frac{e ^{4 x} - 5}{e ^{2 x}}) d x

y^{^{'}} - \frac{5}{x} y = \frac{y ^{5}}{x ^{3}}, y (1) = 1

What is the derivative of ((1+16t^2))/((4t)^3) ?
The derivative of ((1+16t^2))/((4t)^3) is (-16t^2-3)/(64t^4)
What is the first derivative of ((1+16t^2))/((4t)^3) ?
The first derivative of ((1+16t^2))/((4t)^3) is (-16t^2-3)/(64t^4)