What is the derivative of g(t)=(t-sqrt(t))/(t^{1/6)} ?

The derivative of g(t)=(t-sqrt(t))/(t^{1/6)} is (sqrt(t)-1)/(3t^{2/3)}+1/(2t^{1/6)}

What is the first derivative of g(t)=(t-sqrt(t))/(t^{1/6)} ?

The first derivative of g(t)=(t-sqrt(t))/(t^{1/6)} is (sqrt(t)-1)/(3t^{2/3)}+1/(2t^{1/6)}

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derivative of

g (t) = \frac{t - t}{t ^{\frac{1}{6}}}

\frac{t - 1}{3 t ^{\frac{2}{3}}} + \frac{1}{2 t ^{\frac{1}{6}}}

Solution steps

\frac{d}{d t} (\frac{t - t}{t ^{\frac{1}{6}}})

Simplify

\frac{t - t}{t ^{\frac{1}{6}}} : t^{\frac{1}{3}} (t - 1)

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{\frac{1}{3}}, g = t - 1

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

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

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

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

Simplify

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

= \frac{t - 1}{3 t ^{\frac{2}{3}}} + \frac{1}{2 t ^{\frac{1}{6}}}

\frac{d y}{d x} - y = \frac{e ^{x}}{y}

\frac{\partial}{\partial x} ((x^{2})^{- \frac{1}{2}})

u (x) = 3 x^{2} - x + 6

2 cos (2 x), 2 - 2 cos (2 x), [0, \frac{π}{2}]

(6 x + 1)^{2}

What is the derivative of g(t)=(t-sqrt(t))/(t^{1/6)} ?
The derivative of g(t)=(t-sqrt(t))/(t^{1/6)} is (sqrt(t)-1)/(3t^{2/3)}+1/(2t^{1/6)}
What is the first derivative of g(t)=(t-sqrt(t))/(t^{1/6)} ?
The first derivative of g(t)=(t-sqrt(t))/(t^{1/6)} is (sqrt(t)-1)/(3t^{2/3)}+1/(2t^{1/6)}