Logaritmic derivation! 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:
$$y=xe^x\arcsin{x}$$
 A: 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. 
A: 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:
$$
y' = y \cdot \left[ \frac{1}{x} + 1 + \frac{1}{(\arcsin x)\sqrt{1 - x^2}} \right]
= \left[ xe^x\arcsin x \right] \cdot \left[ \frac{1}{x} + 1 + \frac{1}{(\arcsin x)\sqrt{1 - x^2}} \right]
$$
