If #f(x)= cot2 x # and #g(x) = e^(-3x ) #, how do you differentiate #f(g(x)) # using the chain rule?

1 Answer

Answer:

#color(red)(d/dxf(g(x))=6*e^(-3x)csc^2(2*e^(-3x)))#

Explanation:

#f(g(x))=cot(2*g(x))#

#f(g(x))=cot(2*e^(-3x))#

#d/dxf(g(x))=d/dxcot(2*e^(-3x))#

#d/dxf(g(x))=-csc^2(2*e^(-3x))*d/dx(2*e^(-3x))#

#d/dxf(g(x))=-csc^2(2*e^(-3x))*2*d/dx(e^(-3x))#

#d/dxf(g(x))=-csc^2(2*e^(-3x))*2*e^(-3x)*d/dx(-3x)#

#d/dxf(g(x))=-csc^2(2*e^(-3x))*2*e^(-3x)*(-3)#

#color(red)(d/dxf(g(x))=6*e^(-3x)csc^2(2*e^(-3x)))#

God bless....I hope the explanation is useful