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

The d/(dt)(1/(sqrt(1+4t^2))) is -(4t)/((1+4t^2)^{3/2)}

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

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

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

\frac{d}{d t} (1 + 4 t^{2}) = 8 t

= - \frac{1}{2 ( 1 + 4 t ^{2} ) ^{\frac{3}{2}}} \cdot 8 t

Simplify

- \frac{1}{2 ( 1 + 4 t ^{2} ) ^{\frac{3}{2}}} \cdot 8 t : - \frac{4 t}{( 1 + 4 t ^{2} ) ^{\frac{3}{2}}}

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

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\frac{\partial}{\partial x} (3 x^{4} - 65 x^{3})

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