What is the derivative of \sqrt[7]{t}+6sqrt(t^7) ?

The derivative of \sqrt[7]{t}+6sqrt(t^7) is (1+147t^{47/14})/(7t^{6/7)}

What is the first derivative of \sqrt[7]{t}+6sqrt(t^7) ?

The first derivative of \sqrt[7]{t}+6sqrt(t^7) is (1+147t^{47/14})/(7t^{6/7)}

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derivative of

\frac{1 + 147 t ^{\frac{47}{14}}}{7 t ^{\frac{6}{7}}}

Solution steps

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

\frac{d}{d t} (6 t^{7}) = 21 t^{\frac{5}{2}}

= \frac{1}{7 t ^{\frac{6}{7}}} + 21 t^{\frac{5}{2}}

Simplify

\frac{1}{7 t ^{\frac{6}{7}}} + 21 t^{\frac{5}{2}} : \frac{1 + 147 t ^{\frac{47}{14}}}{7 t ^{\frac{6}{7}}}

= \frac{1 + 147 t ^{\frac{47}{14}}}{7 t ^{\frac{6}{7}}}

y = 5 x - x^{2}, y = - x

\frac{\partial}{\partial x} (3 y cos (x))

\frac{1}{s [ 5 + s ]}

y^{^{''}} - 12 y^{^{'}} + 36 y = 0

\frac{d ^{2}}{d x ^{2}} (cos (x))

What is the derivative of \sqrt[7]{t}+6sqrt(t^7) ?
The derivative of \sqrt[7]{t}+6sqrt(t^7) is (1+147t^{47/14})/(7t^{6/7)}
What is the first derivative of \sqrt[7]{t}+6sqrt(t^7) ?
The first derivative of \sqrt[7]{t}+6sqrt(t^7) is (1+147t^{47/14})/(7t^{6/7)}