Determine an equation for the tangent line to the graph of #f(x)=(2x+1)^2+ln(4x-3)# at the point #x=1#?

Please do not skip any steps!

1 Answer
Aug 5, 2017

Answer:

The equation is #y=16x-7#

Explanation:

We need

#(u(x)^n)'=n*u(x)^(n-1)*u'(x)#

#(ln(u(x)))'=1/(u(x))*u'(x)#

#ln1=0#

Our function is

#f(x)=(2x+1)^2+ln(4x-3)#

We calculate the first derivative

#f'(x)=2(2x+1)*2+4/(4x-3)=4(2x+1)+4/(4x-3)#

The equation of the tangent is

#y=f'(a)(x-a)+f(a)#

Here #a=1#

#f(1)=(2+1)^2+ln(4-3)=9#

#f'(1)=4*3+4/1=16#

Therefore, the equation is

#y=16(x-1)+9=16x-16+9=16x-7#

graph{(y-(2x+1)^2-ln(4x-3))(y-16x+7)=0 [-32.2, 8.35, 1.1, 21.38]}