How do you find the derivative of #sin(cosx)#?

1 Answer
Jim G.
Dec 3, 2016

#-sinxcos(cosx)#

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)#

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