What is the derivative of g(t)= 1/(sqrt(t)) ?

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

What is the first derivative of g(t)= 1/(sqrt(t)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

g (t) = \frac{1}{t}

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

Solution steps

\frac{d}{d t} (\frac{1}{t})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= - \frac{1}{2} t^{- \frac{1}{2} - 1}

Simplify

- \frac{1}{2} t^{- \frac{1}{2} - 1} : - \frac{1}{2 t ^{\frac{3}{2}}}

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

\int (x + 1)^{2} e^{- x} d x

\int 4 x d x

x \to 0 - lim (\frac{6 sin ( x )}{x})

\int_{1}^{\infty} \frac{2}{e ^{x}} d x

\frac{\partial}{\partial x} (6 y^{2} - 6 x^{2})

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