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

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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

= 3 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}}

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

Simplify

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

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

\frac{\partial}{\partial x} (6 e^{x})

\int tan^{- 3} (x) sec^{2} (x) sec^{2} (x) d x

x \to 0 lim (\frac{sin ( x ^{8} )}{x})

tan (e^{7 t}) + e^{t a n (7 t)}

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

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