Question #b7835

2 Answers
Surya K.
Jan 22, 2018

#-8/(2x-1)^2#

Explanation:

#d/dx4/(2x-1)#

#4*d/dx1/(2x-1)#

#4*d/dx(2x-1)^-1#

Remember the chain rule: for #(df(u))/dx = (df)/(du)*(du)/(dx)#

Here, #f=u^-1#, and #u=2x-1#

#(df)/(du)=d/(du) u^-1#

Power rule: #d/dxx^n=nx^(n-1)#

#=-1/u^2#

#(du)/dx=d/dx2x-1#

Remember that #d/dxk#, where #k# is a constant, is equal to #0#, and #d/dx2x=2#.

#=2#

So now we have #4(-1/u^2)2#

But remember, #u=2x-1#, so:

#4(-1/(2x-1)^2)2#

#-8/(2x-1)^2#

Jim G.
Jan 22, 2018

#-8/(2x-1)^2#

Explanation:

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

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#4/(2x-1)=4(2x-1)^-1#

#rArrd/dx(4(2x-1)^-1)#

#=-4(2x-1)^-2xxd/dx(2x-1)#

#=-8/(2x-1)^2#

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