What is the derivative of f(x)=ln(ln(x^3e^{-3x})) ?

The derivative of f(x)=ln(ln(x^3e^{-3x})) is (3(-x+1))/(xln(e^{-3x)x^3)}

What is the first derivative of f(x)=ln(ln(x^3e^{-3x})) ?

The first derivative of f(x)=ln(ln(x^3e^{-3x})) is (3(-x+1))/(xln(e^{-3x)x^3)}

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derivative of

f (x) = ln (ln (x^{3} e^{- 3 x}))

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

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{ln ( x ^{3} e ^{- 3 x} )} \cdot \frac{3 ( - x + 1 )}{x}

Simplify

\frac{1}{ln ( x ^{3} e ^{- 3 x} )} \cdot \frac{3 ( - x + 1 )}{x} : \frac{3 ( - x + 1 )}{x ln ( e ^{- 3 x} x ^{3} )}

= \frac{3 ( - x + 1 )}{x ln ( e ^{- 3 x} x ^{3} )}

\frac{\partial}{\partial y} (x^{2} y + y^{2} z + z^{2} x)

\frac{d ^{2}}{d x ^{2}} (\frac{3 x + 5}{2 x - 1})

y^{^{'}} + x y = x y^{- 2}

\frac{d}{d x} (12 x - 3 x^{2})

\frac{1}{1 + 3 x}

What is the derivative of f(x)=ln(ln(x^3e^{-3x})) ?
The derivative of f(x)=ln(ln(x^3e^{-3x})) is (3(-x+1))/(xln(e^{-3x)x^3)}
What is the first derivative of f(x)=ln(ln(x^3e^{-3x})) ?
The first derivative of f(x)=ln(ln(x^3e^{-3x})) is (3(-x+1))/(xln(e^{-3x)x^3)}