Was ist die Ableitung von # cosx ^ 2 #?

Antworten:

#-sin2x#

Erläuterung:

Differentiate using the #color(blue)"chain rule"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(f(g(x)))=f'(g(x))g'(x))color(white)(a/a)|))) ........ (A)#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(cosx)=-sinx)color(white)(a/a)|)))#
#color(blue)"-----------------------------------------------"#

#f(g(x))=cos^2x=(cosx)^2rArrf'(g(x))=2(cosx)^1=2cosx#

and #g(x)=cosxrArrg'(x)=-sinx#
#color(blue)"-----------------------------------------------"#
Substitute these values into (A)

#rArrf'(g(x))=2cosx(-sinx)=-2sinxcosx#

Using the following trig. identity to simplify.

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(sin2x=2sinxcosx)color(white)(a/a)|)))#

#rarrf'(g(x))=-2sinxcosx=-sin2x#

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