What is the d/(dt)((3y^2-t^2)/(y^5)) ?

The d/(dt)((3y^2-t^2)/(y^5)) is (5t^2\frac{dy)/(dt)-2ty-9y^2(dy)/(dt)}{y^6}

What is the first d/(dt)((3y^2-t^2)/(y^5)) ?

The first d/(dt)((3y^2-t^2)/(y^5)) is (5t^2\frac{dy)/(dt)-2ty-9y^2(dy)/(dt)}{y^6}

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\frac{d}{d t} (\frac{3 y ^{2} - t ^{2}}{y ^{5}})

\frac{5 t ^{2} \frac{d y}{d t} - 2 t y - 9 y ^{2} \frac{d y}{d t}}{y ^{6}}

Solution steps

\frac{d}{d t} (\frac{3 y ^{2} - t ^{2}}{y ^{5}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d t} ( 3 y ^{2} - t ^{2} ) y ^{5} - \frac{d}{d t} ( y ^{5} ) ( 3 y ^{2} - t ^{2} )}{( y ^{5} ) ^{2}}

\frac{d}{d t} (3 y^{2} - t^{2}) = 6 y \frac{d y}{d t} - 2 t

\frac{d}{d t} (y^{5}) = 5 y^{4} \frac{d y}{d t}

= \frac{( 6 y \frac{d y}{d t} - 2 t ) y ^{5} - 5 y ^{4} \frac{d y}{d t} ( 3 y ^{2} - t ^{2} )}{( y ^{5} ) ^{2}}

Simplify

\frac{( 6 y \frac{d y}{d t} - 2 t ) y ^{5} - 5 y ^{4} \frac{d y}{d t} ( 3 y ^{2} - t ^{2} )}{( y ^{5} ) ^{2}} : \frac{5 t ^{2} \frac{d y}{d t} - 2 t y - 9 y ^{2} \frac{d y}{d t}}{y ^{6}}

= \frac{5 t ^{2} \frac{d y}{d t} - 2 t y - 9 y ^{2} \frac{d y}{d t}}{y ^{6}}

\frac{d}{d x} (ln (\frac{4}{x x + 1}))

(e^{x^{2} - 1})^{^{'}}

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

\frac{d}{d x} ((x^{3} - 4 x + 1) e^{x})

\frac{d}{d x} (arcsec^{2} (x))

What is the d/(dt)((3y^2-t^2)/(y^5)) ?
The d/(dt)((3y^2-t^2)/(y^5)) is (5t^2\frac{dy)/(dt)-2ty-9y^2(dy)/(dt)}{y^6}
What is the first d/(dt)((3y^2-t^2)/(y^5)) ?
The first d/(dt)((3y^2-t^2)/(y^5)) is (5t^2\frac{dy)/(dt)-2ty-9y^2(dy)/(dt)}{y^6}