009A Sample Final 1, Problem 4

If

$y=x^{2}+\cos(\pi (x^{2}+1))$

compute  ${\frac {dy}{dx}}$  and find the equation for the tangent line at $x_{0}=1$. You may leave your answers in point-slope form.

Foundations:
1. What two pieces of information do you need to write the equation of a line?
You need the slope of the line and a point on the line.
2. What does the Chain Rule state?
For functions $f(x)$  and $g(x),$  $~{\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x).$

Solution:

Step 1:
First, we compute  ${\frac {dy}{dx}}.$ We get
${\frac {dy}{dx}}\,=\,2x-\sin(\pi (x^{2}+1))(2\pi x).$

Step 2:
To find the equation of the tangent line, we first find the slope of the line.
Using $x_{0}=1$  in the formula for  ${\frac {dy}{dx}}$  from Step 1, we get
$m=2(1)-\sin(2\pi )2\pi \,=\,2.$
To get a point on the line, we plug in $x_{0}=1$  into the equation given.
So, we have
$y=1^{2}+\cos(2\pi )=2.$
Thus, the equation of the tangent line is
$y=2(x-1)+2.$

Final Answer:
${\frac {dy}{dx}}=2x-\sin(\pi (x^{2}+1))(2\pi x)$
$y=2(x-1)+2$

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