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.
2. What does the Chain Rule state?
For functions $f(x),g(x)$, $~{\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$

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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