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Why can't I derive this function using normal methods?Text book says that i have to use something called "logarithmic derivation". I don't know if this term exists in English, but that is the direct translation from my language. Apparently you use this method when there is a variable in both the base, and the exponent, but i can't seem to spot this situation in the function that is given to me:

$$y=xe^x\arcsin{x}$$

Eleven-Eleven
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user118751
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  • The English phrase you are probably looking for is "logarithmic differentiation". It's the process where you take the natural logarithm of both sides of the equation and then take $\frac{d}{dx}$ of both sides. – MartianInvader Jan 02 '14 at 19:23

2 Answers2

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Logarithmic differentiation is the English term.

Essentially, you can use the properties of the logarithm to simplify certain functions, and then differentiate.

In your case, however, logarithmic differentiation isn't necessary, just use the product rule.

Consider differentiating $y = x^x$. Taking logarithms, we obtain $$\ln{y} = x\ln{x} $$ which is readily differentiated to obtain $$ \frac{1}{y}\frac{dy}{dx} = \ln{x} + 1 $$ and then you can multiply through by $y$, substitute $y = x^x$, and get the derivative.

Ayesha
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  • Weird. Task specifies that I need to use logarithmic differentiation. Reckon it is a mistake? – user118751 Jan 02 '14 at 19:25
  • Originally, you didn't use LaTeX to specify your function, and then the moderator edited it, so I assume he/she may not have written the function you wanted. – Ayesha Jan 02 '14 at 19:27
  • No, the function is correct even after the edit. We still haven't learned about the product rule involving 3 functions. Only 2 so far. – user118751 Jan 02 '14 at 19:28
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    Two comments: (1) If the task specified logarithmic differentiation, that is because that is what they wanted you to practice. Not because it can't be done otherwise. (2) To differentiate a product of three functions, say $fgh$, apply the two-function product rule to get $(fgh)'=(fg)'h+fgh'$, then compute $(fg)'$ using the product rule. So you don't need to learn a special three factor product rule. – Harald Hanche-Olsen Jan 02 '14 at 19:35
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Logarithmic differentiation here is not necessary. Nevertheless, we can apply it anyways in order to avoid product rule; it's really handy because logarithms transform products and quotients into sums and differences! Indeed, observe that: $$ \ln y = \ln(xe^x\arcsin x) = \ln x + \ln(e^x) + \ln(\arcsin x) = \ln x + x + \ln(\arcsin x) $$ Taking derivatives of both sides, we obtain: $$ \frac{1}{y} \cdot y' = \frac{1}{x} + 1 + \frac{1}{\arcsin x} \cdot \frac{1}{\sqrt{1 - x^2}} $$ Hence, we obtain: $$ y' = y \cdot \left[ \frac{1}{x} + 1 + \frac{1}{(\arcsin x)\sqrt{1 - x^2}} \right] = \left[ xe^x\arcsin x \right] \cdot \left[ \frac{1}{x} + 1 + \frac{1}{(\arcsin x)\sqrt{1 - x^2}} \right] $$

Adriano
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