What is the derivative of 8/(sqrt(64+64t^2)) ?

The derivative of 8/(sqrt(64+64t^2)) is -t/((1+t^2)^{3/2)}

What is the first derivative of 8/(sqrt(64+64t^2)) ?

The first derivative of 8/(sqrt(64+64t^2)) is -t/((1+t^2)^{3/2)}

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derivative of

\frac{8}{64 + 64 t ^{2}}

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

Solution steps

\frac{d}{d t} (\frac{8}{64 + 64 t ^{2}})

Take the constant out:

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

= 8 \frac{d}{d t} (\frac{1}{64 + 64 t ^{2}})

Apply exponent rule:

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

= 8 \frac{d}{d t} ((64 + 64 t^{2})^{- \frac{1}{2}})

Apply the chain rule:

- \frac{1}{2 ( 64 + 64 t ^{2} ) ^{\frac{3}{2}}} \frac{d}{d t} (64 + 64 t^{2})

= - \frac{1}{2 ( 64 + 64 t ^{2} ) ^{\frac{3}{2}}} \frac{d}{d t} (64 + 64 t^{2})

\frac{d}{d t} (64 + 64 t^{2}) = 128 t

= 8 (- \frac{1}{2 ( 64 + 64 t ^{2} ) ^{\frac{3}{2}}} \cdot 128 t)

Simplify

8 (- \frac{1}{2 ( 64 + 64 t ^{2} ) ^{\frac{3}{2}}} \cdot 128 t) : - \frac{t}{( 1 + t ^{2} ) ^{\frac{3}{2}}}

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

\frac{d y}{d x} = 2 cos (x) (y + 1), y (\frac{π}{6}) = 2

x \to 0 + lim ((5 x)^{3 x})

x^{2} y^{^{'}} = 2 x y + y^{2}

\frac{\partial}{\partial x} (2 x^{2} y^{2})

n = 0 \sum \infty 9 (\frac{3}{4})^{n}

What is the derivative of 8/(sqrt(64+64t^2)) ?
The derivative of 8/(sqrt(64+64t^2)) is -t/((1+t^2)^{3/2)}
What is the first derivative of 8/(sqrt(64+64t^2)) ?
The first derivative of 8/(sqrt(64+64t^2)) is -t/((1+t^2)^{3/2)}