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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\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}}}

\frac{d}{d x} (2 x^{4} - 6 x^{3} - 12 x - 8)

n = 5 \sum \infty \frac{1}{n ( n - 1 )}

\int_{2}^{4} (x^{4} - 1) d x

x \to 0 lim (\frac{1 - e ^{x}}{e ^{x}})

x \to 0 lim ((1 + 4 x)^{\frac{6}{x}})

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