Let us look at the function: #y=sin(x)^ln(x)ln(x)^cos(x)#. If we set #u=sin(x)^ln(x)# and #v=ln(x)^cos(x)#, then, by the product rule, #dy/dx=u(dv)/dx+v(du)/dx=sin(x)^ln(x)*(dv)/dx+ln(x)^cos(x)*(du)/dx#.
Thus, we just need to find #(du)/dx# and #(dv)/dx#.
Since these functions involve exponents, let us find what #(ds)/dt# is given #s=f(t)^g(t)#. The power rule does not apply here, since the exponent is non-constant, and the exponent rule does not apply here, as the base is non-constant.
However, we can make the base into a constant variable since #f(t)=e^ln(f(t))#. Thus, #s=e^(ln(f(t))g(t))#. Now, if we use the exponent rule and then the product rule and chain rule, we get #(ds)/dt=((f'(t)g(t))/f(t)+ln(f(t))g'(t))e^(ln(f(t))g(t))=f(t)^g(t)((f'(t)g(t))/f(t)+ln(f(t))g'(t))#.
We know that #u=sin(x)^ln(x)#. Thus, using the formula we found above, we get #(du)/dx=sin(x)^ln(x)((cos(x)ln(x))/(sin(x))+ln(sin(x))/x)#.
Do the same for #v=ln(x)^cos(x)#. You should get #(dv)/dx=ln(x)^cos(x)(cos(x)/(xln(x))-ln(ln(x))sin(x))#.
We said previously that #dy/dx=u(dv)/dx+v(du)/dx=sin(x)^ln(x)*(dv)/dx+ln(x)^cos(x)*(du)/dx#.
Substitute what we got for #(du)/dx# and #(dv)/dx# to get the final answer:
#dy/dx=sin(x)^ln(x)ln(x)^cos(x)(cos(x)/(xln(x))-ln(ln(x))sin(x))+ln(x)^cos(x)sin(x)^ln(x)((cos(x)ln(x))/(sin(x))+ln(sin(x))/x)#.
If you want, you can try to further simplify this.
