How do you find the derivative of #y=x^cos(x)#?
1 Answer
#y'=x^cos(x)*((cos(x))/x-sin(x)ln(x))# Solution :
#y=f(x)^g(x)# , this type of questions usually solve by following two methods,Explanation (I)
taking
#ln# of both sides, we get,
#ln(y)=g(x)*ln(f(x))# Now differentiating both sides with respect to
#x# using Product Rule
#1/y*y'=g(x)*(f'(x))/f(x)+ln(f(x))*g'(x)# ,
#y'=y(g(x)*(f'(x))/f(x)+ln(f(x))*g'(x))# ,
#y'=f(x)^g(x)(g(x)*(f'(x))/f(x)+ln(f(x))*g'(x))# ,Similarly following for
#y=x^cos(x)# , we get
#lny=cos(x)*ln(x)#
#(y')/y=(cos(x))/x+ln(x)(-sin(x))#
#y'=y*((cos(x))/x-sin(x)ln(x))#
#y'=x^cos(x)*((cos(x))/x-sin(x)ln(x))# Explanation (II)
#y=f(x)^g(x)# first do the differentiation keeping
#g(x)# as constant and f(x) as it is (i.e.#x^n# ), then#f(x)# as constant and g(x) as it is (i.e.#a^x# ), this is quick way to do these type of questions,like,
#y'=g(x)(f(x))^(g(x)-1)*f'(x)+f(x)^g(x)*ln(f(x))*g'(x)# following in the same way,
#y'=cos(x)(x^(cos(x)-1))+x^cos(x)(lnx)(-sin(x))#
#y'=x^cos(x)(cos(x)(x^-1)+(lnx)(-sin(x)))#
#y'=x^cos(x)(cos(x)/x-sin(x)lnx)#
