How do you find the derivative of #sin(cosx)#?
1 Answer
Dec 3, 2016
Explanation:
differentiate using the
#color(blue)"chain rule"#
#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(2/2)|)))to(A)#
#"let " u=cosxrArr(du)/(dx)=-sinx#
#"and " y=sinurArr(dy)/(du)=cosu# Substitute into ( A), changing u back to terms of x.
#rArrdy/dx=cosuxx(-sinx)=-sinxcos(cosx)#