Difference between revisions of "009A Sample Final 1, Problem 4"

Revision as of 15:07, 22 February 2016

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:

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

@@ Line 18: / Line 18: @@
 |First, we compute <math>\frac{dy}{dx}</math>. We get
 |-
-|<math>\frac{dy}{dx}=2x-\sin(\pi(x^2+1))(2\pi x)</math>.
+|
+::<math>\frac{dy}{dx}=2x-\sin(\pi(x^2+1))(2\pi x)</math>.
 |}
@@ Line 26: / Line 27: @@
 |To find the equation of the tangent line, we first find the slope of the line.
 |-
-|Using <math>x_0=1</math> in the formula for <math>\frac{dy}{dx}</math> from Step 1, we get
+|Using <math style="vertical-align: -3px">x_0=1</math> in the formula for <math style="vertical-align: -12px">\frac{dy}{dx}</math> from Step 1, we get
 |-
-|<math>m=2(1)-\sin(2\pi)2\pi=2</math>.
+|
+::<math>m=2(1)-\sin(2\pi)2\pi=2</math>.
 |-
-|To get a point on the line, we plug in <math>x_0=1</math> into the equation given.
+|To get a point on the line, we plug in <math style="vertical-align: -3px">x_0=1</math> into the equation given.
 |-
-|So, we have <math>y=1^2+\cos(2\pi)=2</math>.
+|So, we have <math style="vertical-align: -5px">y=1^2+\cos(2\pi)=2</math>.
 |-
-|Thus, the equation of the tangent line is <math>y=2(x-1)+2</math>.
+|Thus, the equation of the tangent line is <math style="vertical-align: -5px">y=2(x-1)+2</math>.
 |}