What is the derivative of g(t)=(t-sqrt(t))/(t^{1/9)} ?

The derivative of g(t)=(t-sqrt(t))/(t^{1/9)} is (7(sqrt(t)-1))/(18t^{11/18)}+1/(2t^{1/9)}

What is the first derivative of g(t)=(t-sqrt(t))/(t^{1/9)} ?

The first derivative of g(t)=(t-sqrt(t))/(t^{1/9)} is (7(sqrt(t)-1))/(18t^{11/18)}+1/(2t^{1/9)}

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derivative of

g (t) = \frac{t - t}{t ^{\frac{1}{9}}}

\frac{7 ( t - 1 )}{18 t ^{\frac{11}{18}}} + \frac{1}{2 t ^{\frac{1}{9}}}

Solution steps

\frac{d}{d t} (\frac{t - t}{t ^{\frac{1}{9}}})

Simplify

\frac{t - t}{t ^{\frac{1}{9}}} : t^{\frac{7}{18}} (t - 1)

= \frac{d}{d t} (t^{\frac{7}{18}} (t - 1))

Apply the Product Rule:

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

f = t^{\frac{7}{18}}, g = t - 1

= \frac{d}{d t} (t^{\frac{7}{18}}) (t - 1) + \frac{d}{d t} (t - 1) t^{\frac{7}{18}}

\frac{d}{d t} (t^{\frac{7}{18}}) = \frac{7}{18 t ^{\frac{11}{18}}}

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

= \frac{7}{18 t ^{\frac{11}{18}}} (t - 1) + \frac{1}{2 t ^{\frac{1}{2}}} t^{\frac{7}{18}}

Simplify

\frac{7}{18 t ^{\frac{11}{18}}} (t - 1) + \frac{1}{2 t ^{\frac{1}{2}}} t^{\frac{7}{18}} : \frac{7 ( t - 1 )}{18 t ^{\frac{11}{18}}} + \frac{1}{2 t ^{\frac{1}{9}}}

= \frac{7 ( t - 1 )}{18 t ^{\frac{11}{18}}} + \frac{1}{2 t ^{\frac{1}{9}}}

x^{2} y^{^{''}} + 7 x y^{^{'}} + 5 y = 0

\frac{d}{d x} (\frac{2 + 3 x}{x ^{2}})

\int sin^{18} (x) cos^{3} (x) d x

g (x) = 4 x^{2} (x^{3} + 5 x)

\int (- 8 sec (x) tan (x) + 5 sec^{2} (x)) d x

What is the derivative of g(t)=(t-sqrt(t))/(t^{1/9)} ?
The derivative of g(t)=(t-sqrt(t))/(t^{1/9)} is (7(sqrt(t)-1))/(18t^{11/18)}+1/(2t^{1/9)}
What is the first derivative of g(t)=(t-sqrt(t))/(t^{1/9)} ?
The first derivative of g(t)=(t-sqrt(t))/(t^{1/9)} is (7(sqrt(t)-1))/(18t^{11/18)}+1/(2t^{1/9)}