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How would you simplify something like $$\frac{d(1+5x)}{d \ln{x}}$$

such that the denominator ends up just being $dx$. Presumably you can't just do

$$\frac{d(1+5e^x)}{dx}$$

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5 Answers 5

The beautiful thing about Leibnitz' notation is that you can often get away with treating it like it's a fraction. First $d(1+5x) = 5 \ dx$ while $d \ln x = \frac{1}{x} \ dx$. Thus:

$$\frac{d(1+5x)}{d \ln x} = \frac{5 \ dx}{\frac{1}{x} \ dx} = 5x \frac{dx}{dx} = 5x \, . $$

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I would define $\ln x$ as $y$ and $x$ as $\text{e}^y$ and then modify the other side of the equality accordingly.

If you include the rest of the formula, I can help more.

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Love too much your attempts. –  Babak S. Jan 11 '13 at 20:41

Essentially, you want to differentiate the following w.r.t. $\ln x$: $$(1+5x)$$

So, let $u = \ln x$.

$$x = e^{u}$$

$$1+5x = 1 + 5e^u$$ Now, differentiate the R.H.S. w.r.t. u. $$\frac{d}{d \ln x}\left[1+5x\right] = \frac{d}{du}\left[1+5e^u\right]$$ $$\frac{d}{d \ln x}\left[1+5x\right] = 5e^u$$ $$\frac{d}{d \ln x}\left[1+5x\right] = 5x$$

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$$\frac{d(1+5x)}{d \ln x} = \frac{d(1+5x)}{dx}\frac{dx}{d\ln x}=5\left(\frac{d\ln x}{dx}\right)^{-1}=5\left(\frac{1}{x}\right)^{-1}=5x$$

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Well, you actually can, though I wouldn't re-use the same variable $x$. In calculating $\frac{d(1 + 5x)}{d \ln{x}}$, you can first use the substitution $u = \ln{x}$, in which case our derivative becomes $\frac{d(1 + 5e^u)}{d u} = 5e^u = 5x$.

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