If #f(x)= cot5 x # and #g(x) = sqrt(-x+3 #, how do you differentiate #f(g(x)) # using the chain rule?

1 Answer
Mar 16, 2017

Answer:

#(df)/(dx)=-(csc^2(sqrt(-x+3)))/(2sqrt(-x+3))#

Explanation:

As #f(x)=cot5x# and #g(x)=sqrt(-x+3)#, therefore

#f(g(x))=cot(5sqrt(-x+3))#

According to chain rule when #f=f(g(x))#,

#(df)/(dx)=(df)/(dg)xx(dg)/(dx)#

As #f=cot(g(x))#,

#(df)/(dg)=-csc^2(g(x)=-csc^2(sqrt(-x+3))#

and as #g(x)=sqrt(-x+3)#

#(dg)/(dx)=1/(2sqrt(-x+3))xxd/(dx)(-x+3)=-1/(2sqrt(-x+3))#

(note that we have again applied chain rule here as #-x+3# is again a function of #x# and hence we should multiply by #d/(dx)(-x+3)#)

Hence #(df)/(dx)=-csc^2(sqrt(-x+3))xx-1/(2sqrt(-x+3))#

= #-(csc^2(sqrt(-x+3)))/(2sqrt(-x+3))#