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!

Aug 5, 2017

The equation is $y = 16 x - 7$

Explanation:

We need

$\left(u {\left(x\right)}^{n}\right) ' = n \cdot u {\left(x\right)}^{n - 1} \cdot u ' \left(x\right)$

$\left(\ln \left(u \left(x\right)\right)\right) ' = \frac{1}{u \left(x\right)} \cdot u ' \left(x\right)$

$\ln 1 = 0$

Our function is

$f \left(x\right) = {\left(2 x + 1\right)}^{2} + \ln \left(4 x - 3\right)$

We calculate the first derivative

$f ' \left(x\right) = 2 \left(2 x + 1\right) \cdot 2 + \frac{4}{4 x - 3} = 4 \left(2 x + 1\right) + \frac{4}{4 x - 3}$

The equation of the tangent is

$y = f ' \left(a\right) \left(x - a\right) + f \left(a\right)$

Here $a = 1$

$f \left(1\right) = {\left(2 + 1\right)}^{2} + \ln \left(4 - 3\right) = 9$

$f ' \left(1\right) = 4 \cdot 3 + \frac{4}{1} = 16$

Therefore, the equation is

$y = 16 \left(x - 1\right) + 9 = 16 x - 16 + 9 = 16 x - 7$

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