What is the derivative of 7/(\sqrt[3]{3t^2+t)} ?

The derivative of 7/(\sqrt[3]{3t^2+t)} is -(7(6t+1))/(3(3t^2+t)^{4/3)}

What is the first derivative of 7/(\sqrt[3]{3t^2+t)} ?

The first derivative of 7/(\sqrt[3]{3t^2+t)} is -(7(6t+1))/(3(3t^2+t)^{4/3)}

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derivative of

- \frac{7 ( 6 t + 1 )}{3 ( 3 t ^{2} + t ) ^{\frac{4}{3}}}

Solution steps

Take the constant out:

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

Apply exponent rule:

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

= 7 \frac{d}{d t} ((3 t^{2} + t)^{- \frac{1}{3}})

Apply the chain rule:

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

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

\frac{d}{d t} (3 t^{2} + t) = 6 t + 1

= 7 (- \frac{1}{3 ( 3 t ^{2} + t ) ^{\frac{4}{3}}} (6 t + 1))

Simplify

7 (- \frac{1}{3 ( 3 t ^{2} + t ) ^{\frac{4}{3}}} (6 t + 1)) : - \frac{7 ( 6 t + 1 )}{3 ( 3 t ^{2} + t ) ^{\frac{4}{3}}}

= - \frac{7 ( 6 t + 1 )}{3 ( 3 t ^{2} + t ) ^{\frac{4}{3}}}

\frac{d}{d x} ((ln (x^{5}))^{2})

f (x) = \frac{20}{3} x^{3} + 380 x^{2}

\frac{\partial}{\partial x} (3 ln (x) + ln (y))

\frac{\partial}{\partial x} (400 e^{x y})

n = 1 \sum \infty n sin (n)

What is the derivative of 7/(\sqrt[3]{3t^2+t)} ?
The derivative of 7/(\sqrt[3]{3t^2+t)} is -(7(6t+1))/(3(3t^2+t)^{4/3)}
What is the first derivative of 7/(\sqrt[3]{3t^2+t)} ?
The first derivative of 7/(\sqrt[3]{3t^2+t)} is -(7(6t+1))/(3(3t^2+t)^{4/3)}