What is (\partial)/(\partial y)((t^2)/(4y^4)) ?

The answer to (\partial)/(\partial y)((t^2)/(4y^4)) is -(t^2)/(y^5)

Popular >

(\partial)/(\partial y)((t^2)/(4y^4))

\frac{\partial}{\partial y} (\frac{t ^{2}}{4 y ^{4}})

- \frac{t ^{2}}{y ^{5}}

Solution steps

\frac{\partial}{\partial y} (\frac{t ^{2}}{4 y ^{4}})

Treat

t

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{t ^{2}}{4} \frac{\partial}{\partial y} (\frac{1}{y ^{4}})

Apply exponent rule:

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

= \frac{t ^{2}}{4} \frac{\partial}{\partial y} (y^{- 4})

Apply the Power Rule:

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

= \frac{t ^{2}}{4} (- 4 y^{- 4 - 1})

Simplify

\frac{t ^{2}}{4} (- 4 y^{- 4 - 1}) : - \frac{t ^{2}}{y ^{5}}

= - \frac{t ^{2}}{y ^{5}}

\int \frac{3 x + 1}{( x ^{2} + 2 ) ( x + 3 )} d x

\int 2 x x - 3 d x

\int \frac{1}{( y - 3 ) ( y - 2 ) ( y - 1 )} d y

\int \frac{e ^{x} + e ^{- x}}{e ^{x} - e ^{- x}} d x

x^{2} y^{^{''}} + 2 x y^{^{'}} - 12 y = 0, y (1) = 0, y^{^{'}} (1) = - 2

What is (\partial)/(\partial y)((t^2)/(4y^4)) ?
The answer to (\partial)/(\partial y)((t^2)/(4y^4)) is -(t^2)/(y^5)