# Derivatives and tangent lines

• Jul 27th 2009, 07:54 PM
goldenroll
Derivatives and tangent lines
I'm having some trouble on these questions. On other questions for this I was using derivatives to get the answers but on these I don't know how to apply it. Can someone explain how to use derivatives to find these answers?

1. Find the equation of the tangent line to the curve given by y=2xe^x. At (0,0).

2. Find the exact points (x,y) on the curve y=x ln x where the tangent line is horizontal.

3. Find the exact point (x,y) on the curve y= sin^2 x, 0<x<pie where the tangent is horizontal.
• Jul 27th 2009, 08:06 PM
VonNemo19
Quote:

Originally Posted by goldenroll
I'm having some trouble on these questions. On other questions for this I was using derivatives to get the answers but on these I don't know how to apply it. Can someone explain how to use derivatives to find these answers?

1. Find the equation of the tangent line to the curve given by y=2xe^x. At (0,0).

2. Find the exact points (x,y) on the curve y=x ln x where the tangent line is horizontal.

3. Find the exact point (x,y) on the curve y= sin^2 x, 0<x<pie where the tangent is horizontal.

1. $y'=2xe^x+2e^x\Rightarrow{m}=2$ at $x=0$.
Then the line tangent at (0,0) is given by $(y-0)=2(x-0)$.

2. Horizontal $\Rightarrow{y'}=0$. Then $y'=1+\ln{x}=0\Rightarrow{x}=\frac{1}{e}$

Write the line in the same manner as the first problem I laid out.

3. I leave this to you
• Jul 27th 2009, 08:10 PM
HallsofIvy
Quote:

Originally Posted by goldenroll
I'm having some trouble on these questions. On other questions for this I was using derivatives to get the answers but on these I don't know how to apply it. Can someone explain how to use derivatives to find these answers?

1. Find the equation of the tangent line to the curve given by y=2xe^x. At (0,0).

The derivative is the slope of the tangent line. The tangent line at (0,0) is y= m(x- 0)+ 0= mx where m is the derivative. What is the derivative of $y= 2xe^x$? What is that when x= 0?

Quote:

2. Find the exact points (x,y) on the curve y=x ln x where the tangent line is horizontal.
A line is horizontal when its slope is 0. What is the derivative of x ln x? Set that equal to 0 and solve for x.

Quote:

3. Find the exact point (x,y) on the curve y= sin^2 x, 0<x<pie where the tangent is horizontal.
Same as (2).
• Jul 27th 2009, 08:25 PM
goldenroll
Oh so on the first one am i using the chain rule? since its 2xe^x + 2e^x and then I apply the formula for finding a equation of a line.
• Jul 27th 2009, 08:41 PM
VonNemo19
Quote:

Originally Posted by goldenroll
Oh so on the first one am i using the chain rule? since its 2xe^x + 2e^x and then I apply the formula for finding a equation of a line.

No, not quite. You use the product rule. IE: $\frac{d}{dx}[u\cdot{v}]=u\cdot{v'}+v\cdot{u'}$

In this case, $u=2x$, and $v=e^x$