Question

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!

Answer 1

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]}