What is the d/(dy)((t^2)/(4y^4)) ?

The d/(dy)((t^2)/(4y^4)) is -(t^2)/(y^5)

What is the first d/(dy)((t^2)/(4y^4)) ?

The first d/(dy)((t^2)/(4y^4)) is -(t^2)/(y^5)

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

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

Solution steps

\frac{d}{d 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{d}{d y} (\frac{1}{y ^{4}})

Apply exponent rule:

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

= \frac{t ^{2}}{4} \frac{d}{d 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}}

\frac{d}{d x} (10 x e^{x^{2}})

\frac{d}{d x} (4^{\frac{x}{2}})

f (x) = 12 x

x^{- 7 c o s (x)}

\frac{1}{2 + 3 x}

What is the d/(dy)((t^2)/(4y^4)) ?
The d/(dy)((t^2)/(4y^4)) is -(t^2)/(y^5)
What is the first d/(dy)((t^2)/(4y^4)) ?
The first d/(dy)((t^2)/(4y^4)) is -(t^2)/(y^5)