Y= x^(ax^2 ) dgn x>0, maka dy/dx adalah ..
Takamori37
Hati-hati jebakan dalam mendiferensiasikan x berpangkat f(x)
![$\begin{align}y&=x^{ax^2}=e^{ax^2\ln x} \\ \frac{dy}{dx}&=e^{ax^2\ln x}(ax^2\ln x)' \\ &=x^{ax^2}\times (2ax\ln x+ax^2.\frac1x) \\ &=x^{ax^2}\times(2ax\ln x+ax)\ \ [B]\end{align}](https://tex.z-dn.net/?f=%24%5Cbegin%7Balign%7Dy%26%3Dx%5E%7Bax%5E2%7D%3De%5E%7Bax%5E2%5Cln+x%7D+%5C%5C+%5Cfrac%7Bdy%7D%7Bdx%7D%26%3De%5E%7Bax%5E2%5Cln+x%7D%28ax%5E2%5Cln+x%29%27+%5C%5C+%26%3Dx%5E%7Bax%5E2%7D%5Ctimes+%282ax%5Cln+x%2Bax%5E2.%5Cfrac1x%29+%5C%5C+%26%3Dx%5E%7Bax%5E2%7D%5Ctimes%282ax%5Cln+x%2Bax%29%5C+%5C+%5BB%5D%5Cend%7Balign%7D "$\begin{align}y&=x^{ax^2}=e^{ax^2\ln x} \\ \frac{dy}{dx}&=e^{ax^2\ln x}(ax^2\ln x)' \\ &=x^{ax^2}\times (2ax\ln x+ax^2.\frac1x) \\ &=x^{ax^2}\times(2ax\ln x+ax)\ \ [B]\end{align}")
1 votes
Thanks 0
puccii
makasih atas jawabannya
puccii
tapi jwaban diatas gk ada dipilihannya
Takamori37
Ada apa saja ?
puccii
a. x^(ax^2 ) (2ax log x +a)
b. x^(ax^2 ) (2ax log x +ax)
c. x^(ax^2 ) (2ax log x +2ax)
d. x^(ax^2 ) (2ax log (ax) +ax)
e. x^(ax^2 ) (2ax log (ax) +2ax)
Takamori37
Akan diralat dahulu.
Takamori37
Silakan direfresh
puccii
oke, makasih yaa
Takamori37
Iya ~
fatihahalhusna
1 votes
Thanks 0
More Questions From This User See All