Question #b7835

2 Answers
Jan 22, 2018

-8/(2x-1)^28(2x1)2

Explanation:

d/dx4/(2x-1)ddx42x1

4*d/dx1/(2x-1)4ddx12x1

4*d/dx(2x-1)^-14ddx(2x1)1

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

Here, f=u^-1f=u1, and u=2x-1u=2x1

(df)/(du)=d/(du) u^-1dfdu=dduu1

Power rule: d/dxx^n=nx^(n-1)ddxxn=nxn1

=-1/u^2=1u2

(du)/dx=d/dx2x-1dudx=ddx2x1

Remember that d/dxkddxk, where kk is a constant, is equal to 00, and d/dx2x=2ddx2x=2.

=2=2

So now we have 4(-1/u^2)24(1u2)2

But remember, u=2x-1u=2x1, so:

4(-1/(2x-1)^2)24(1(2x1)2)2

-8/(2x-1)^28(2x1)2

Jan 22, 2018

-8/(2x-1)^28(2x1)2

Explanation:

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

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