What is e^{-ln(1+x)}*derivative of e^{ln(1+x}-1) ?

The solution to e^{-ln(1+x)}*derivative of e^{ln(1+x}-1) is (\frac{dx)/(dx)}{x+1}

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simplify e^{-ln(1+x)}*derivative of e^{ln(1+x}-1)

simplify

e^{- l n (1 + x)} \cdot \frac{d}{d x} (e^{l n (1 + x)} - 1)

\frac{\frac{d x}{d x}}{x + 1}

Solution steps

e^{- l n (1 + x)} \frac{d}{d x} (e^{l n (1 + x)} - 1)

Apply exponent rule:

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

= \frac{1}{e ^{l n (x + 1)}} \frac{d}{d x} (e^{l n (x + 1)} - 1)

\frac{1}{e ^{l n (1 + x)}} = \frac{1}{1 + x}

= \frac{1}{x + 1} \frac{d}{d x} (e^{l n (x + 1)} - 1)

Apply log rule:

a^{l o g_{a} (b)} = b

= \frac{1}{x + 1} \frac{d}{d x} (x + 1 - 1)

Simplify

= \frac{\frac{d x}{d x}}{x + 1}

e^{- t} cos (b t)

c \cdot e^{- 3 x}

(\frac{8}{0.52}) \cdot (0.11 8^{2} - 0.07 8^{2})

e^{- \frac{0.6 π}{1 - 0. 6 ^{2}}}

What is e^{-ln(1+x)}*derivative of e^{ln(1+x}-1) ?
The solution to e^{-ln(1+x)}*derivative of e^{ln(1+x}-1) is (\frac{dx)/(dx)}{x+1}