Question 201612
(Because of some limitations of the software on the Algebra.com site, I am doing to use dy/dx instead of y'.)
Let {{{u = 4x^2 -8x + 3}}} then {{{(du)/(dx) = 8x - 8}}}
Substituting u into your equation we get
{{{y = u^4}}}
And, using the chain rule, we get
{{{(dy)/(dx) = 4*u^3*((du)/(dx))}}}
Substituting back in for both u and du/dx we get:
{{{(dy)/(dx) = 4(4x^2 -8x + 3)^3*(8x - 8)}}}<br>
To find the equation of a  tangent line (or any line) we need, in general, a point on the line and the slope for the line. We have the point, (2, 81). We need the slope. But y' gives us a formula for finding the slope of the tangent. We can use it to find the slope at x=2:
{{{(dy)/(dx) = 4(4(2)^2 -8(2) + 3)^3*(8(2) -8)}}}
Squaring
{{{(dy)/(dx) = 4(4(4) -8(2) + 3)^3*(8(2) -8)}}}
Multiplying inside both parentheses
{{{(dy)/(dx) = 4(16 -16 + 3)^3*(16 -8)}}}
Adding and subtracting inside both parentheses
{{{(dy)/(dx) = 4(3)^3*(8)}}}
Cubing 3
{{{(dy)/(dx) = 4(27)*(8)}}}
Multiplying
{{{(dy)/(dx) = 864}}}<br>
Now we can use either the point-slope form {{{y - y[1] = m*(x - x[1])}}} or the slope intercept form {{{y = mx + b}}} to write the equation of the line through (2, 81) with a slope of 864. Using the point slope form we get:
{{{y - 81 = 864(x - 2)}}}
If you need to have your answer in slope-intercept form, multiply
{{{y - 81 = 864x - 1728)}}}
Then add 81
{{{y = 864x - 1647}}}
or
{{{y = 864x + (-1647)}}}