What is the d/(dy)(9xe^{-1/(y^3)}) ?

The d/(dy)(9xe^{-1/(y^3)}) is (27e^{-1/(y^3)}x)/(y^4)

What is the first d/(dy)(9xe^{-1/(y^3)}) ?

The first d/(dy)(9xe^{-1/(y^3)}) is (27e^{-1/(y^3)}x)/(y^4)

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\frac{d}{d y} (9 x e^{- \frac{1}{y ^{3}}})

\frac{27 e ^{- \frac{1}{y ^{3}}} x}{y ^{4}}

Solution steps

\frac{d}{d y} (9 x e^{- \frac{1}{y ^{3}}})

Treat

x

as a constant

Take the constant out:

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

= 9 x \frac{d}{d y} (e^{- \frac{1}{y ^{3}}})

Apply the chain rule:

e^{- \frac{1}{y ^{3}}} \frac{d}{d y} (- \frac{1}{y ^{3}})

= e^{- \frac{1}{y ^{3}}} \frac{d}{d y} (- \frac{1}{y ^{3}})

\frac{d}{d y} (- \frac{1}{y ^{3}}) = \frac{3}{y ^{4}}

= 9 x e^{- \frac{1}{y ^{3}}} \frac{3}{y ^{4}}

Simplify

9 x e^{- \frac{1}{y ^{3}}} \frac{3}{y ^{4}} : \frac{27 e ^{- \frac{1}{y ^{3}}} x}{y ^{4}}

= \frac{27 e ^{- \frac{1}{y ^{3}}} x}{y ^{4}}

\int e^{x^{2} + 3} d x

y = 8 x - x^{2}, (1, 7)

y^{^{''}} - 2 y^{^{'}} - 8 y = 0 y (0) = - 2, y^{^{'}} (0) = 4

\int \frac{t + 2 t ^{2}}{t} d t

\frac{x d y}{d x} = 4 y

What is the d/(dy)(9xe^{-1/(y^3)}) ?
The d/(dy)(9xe^{-1/(y^3)}) is (27e^{-1/(y^3)}x)/(y^4)
What is the first d/(dy)(9xe^{-1/(y^3)}) ?
The first d/(dy)(9xe^{-1/(y^3)}) is (27e^{-1/(y^3)}x)/(y^4)