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 ), thenf(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)
