What is the derivative of g(t)=(t-sqrt(t))/(t^{1/3)} ?

The derivative of g(t)=(t-sqrt(t))/(t^{1/3)} is (sqrt(t)-1)/(6t^{5/6)}+1/(2t^{1/3)}

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

The first derivative of g(t)=(t-sqrt(t))/(t^{1/3)} is (sqrt(t)-1)/(6t^{5/6)}+1/(2t^{1/3)}

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derivative of

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

\frac{t - 1}{6 t ^{\frac{5}{6}}} + \frac{1}{2 t ^{\frac{1}{3}}}

Solution steps

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

Simplify

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

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

Apply the Product Rule:

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

f = t^{\frac{1}{6}}, g = t - 1

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

\frac{d}{d t} (t^{\frac{1}{6}}) = \frac{1}{6 t ^{\frac{5}{6}}}

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

= \frac{1}{6 t ^{\frac{5}{6}}} (t - 1) + \frac{1}{2 t ^{\frac{1}{2}}} t^{\frac{1}{6}}

Simplify

\frac{1}{6 t ^{\frac{5}{6}}} (t - 1) + \frac{1}{2 t ^{\frac{1}{2}}} t^{\frac{1}{6}} : \frac{t - 1}{6 t ^{\frac{5}{6}}} + \frac{1}{2 t ^{\frac{1}{3}}}

= \frac{t - 1}{6 t ^{\frac{5}{6}}} + \frac{1}{2 t ^{\frac{1}{3}}}

\int \frac{e ^{x}}{e ^{x} + e ^{- x}} d x

n = 0 \sum \infty \frac{1}{2 ^{2^{n}}}

y = π^{3 x + 1}

\frac{d z}{d z}

\frac{d}{d x} ((x + y + 1)^{2})

What is the derivative of g(t)=(t-sqrt(t))/(t^{1/3)} ?
The derivative of g(t)=(t-sqrt(t))/(t^{1/3)} is (sqrt(t)-1)/(6t^{5/6)}+1/(2t^{1/3)}
What is the first derivative of g(t)=(t-sqrt(t))/(t^{1/3)} ?
The first derivative of g(t)=(t-sqrt(t))/(t^{1/3)} is (sqrt(t)-1)/(6t^{5/6)}+1/(2t^{1/3)}