how do i differentiate this:
y=sin(x^sinx)
so far i've manage to do the following:
ln(y)=ln(sin(x^sinx))
but then i reach
$\displaystyle \frac{1}{y} \frac{dy}{dx }=? $
and im not sure how to differentiate the RHS
Thanks!
try the following:
CHAIN RULE: y=sin(x^sin(x)) y'=cos(x^sin(x))*x^sin(x)' = (I); now we forget the rest and only look at x^sin(x)
y=x^sin(x)
ln(y)=ln(x^sin(x))=sin(x)*ln(x) (as ln(a^b)=b*ln(a))
Now we do some differntiating
ln(y)'=y'/y, sin(x)*ln(x)'=sin(x)'*ln(x)+sin(x)*ln(x)'=cos(x)ln (x)+sin(x)/x
->y'/y=cos(x)ln(x)+sin(x)/x
Now the interesting part. We are not looking for ln(y)' but (as f(x)=y) we are looking for y', which we can find in the equation above. Hence, u solve the equation for y'
y'=y(cos(x)ln(x)+sin(x)/x)=x'sin(x)*(cos(x)ln(x)+sin(x)/x) (II)
Now plug (II) in in (I) and you have the solution
(x^sin(x)*cos(x^(sin(x))*(cos(x)ln(x)+sin(x))/x
Generally the point of logarithmic differentiation is the following:
f(x)=y=a^x
-> ln(y)=a*ln(x)
-> ln(y)'=(a*ln(x))'
-> y'/y = (a'ln(x) + a/x)
->y'=y*(a'ln(x) + a/x)
The easiest way to apply logarithmic differentiation is to a function in the form of f(x)=a^g(x), thus in your case x^sin(x). Henceforth you solve the solution until the only thing left is to differntiate x^sin(x), which is done in (I). sin(x^sin(x))= cos(x^sin(x))*x^sin(x)' = cos(x^sin(x)*(II))
The way how to differntiate x^sin(x) is shown in my previous thread, whereas the solution to (x^sin(x))' equals (II).
Thus the soultion to the whole problem is cos(x^sin(x)*x^sin(x)*(cos(x)ln(x)+(sin(x)/x))