What is the derivative of y= 1/t+1/(t^2)-1/(sqrt(t)) ?

The derivative of y= 1/t+1/(t^2)-1/(sqrt(t)) is (-2t^{5/2}-4t^{3/2}+t^3)/(2t^{9/2)}

What is the first derivative of y= 1/t+1/(t^2)-1/(sqrt(t)) ?

The first derivative of y= 1/t+1/(t^2)-1/(sqrt(t)) is (-2t^{5/2}-4t^{3/2}+t^3)/(2t^{9/2)}

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derivative of

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

\frac{- 2 t ^{\frac{5}{2}} - 4 t ^{\frac{3}{2}} + t ^{3}}{2 t ^{\frac{9}{2}}}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

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

Simplify

- \frac{1}{t ^{2}} - \frac{2}{t ^{3}} - (- \frac{1}{2 t ^{\frac{3}{2}}}) : \frac{- 2 t ^{\frac{5}{2}} - 4 t ^{\frac{3}{2}} + t ^{3}}{2 t ^{\frac{9}{2}}}

= \frac{- 2 t ^{\frac{5}{2}} - 4 t ^{\frac{3}{2}} + t ^{3}}{2 t ^{\frac{9}{2}}}

(x^{2} + a^{2})^{e^{x^{2}}}

\frac{d}{d x} (2 a - b x)

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

\int x^{2} + x + 1 d x

\frac{\partial}{\partial x} ((y + x)^{2})

What is the derivative of y= 1/t+1/(t^2)-1/(sqrt(t)) ?
The derivative of y= 1/t+1/(t^2)-1/(sqrt(t)) is (-2t^{5/2}-4t^{3/2}+t^3)/(2t^{9/2)}
What is the first derivative of y= 1/t+1/(t^2)-1/(sqrt(t)) ?
The first derivative of y= 1/t+1/(t^2)-1/(sqrt(t)) is (-2t^{5/2}-4t^{3/2}+t^3)/(2t^{9/2)}