# Chain Rule Derivative (Finding the equation for the tangent line)

• Oct 22nd 2007, 03:33 PM
FalconPUNCH!
Chain Rule Derivative (Finding the equation for the tangent line)
The equation is $y = x^2 e^{-x}$ with the point $(1, 1/e)$

I used the chain rule (I hope I used it correctly)

$f(x) = x^2$ , $f'(x) = 2x$
$g(x) = e^{-x}$ , $g'(x) = e^{-x}$

$y' = 2(e^{-x}) e^{-x}$

The problem is I don't know how to find the equation for the tangent line to the curve at the given point.
• Oct 22nd 2007, 04:19 PM
galactus
That's the product rule, not the chain rule.

$\frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}$

If x=1, then the slope at that point is $\frac{1}{e}$

Now, use your given points in y=mx+b to find the line equation.
• Oct 22nd 2007, 05:53 PM
FalconPUNCH!
Quote:

Originally Posted by galactus
That's the product rule, not the chain rule.

$\frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}$

If x=1, then the slope at that point is $\frac{1}{e}$

Now, use your given points in y=mx+b to find the line equation.

I don't understand how you got that equation. Can you explain it to me? Sorry if it's asking too much.
• Oct 22nd 2007, 06:51 PM
topsquark
Quote:

Originally Posted by galactus
That's the product rule, not the chain rule.

$\frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}$

If x=1, then the slope at that point is $\frac{1}{e}$

Now, use your given points in y=mx+b to find the line equation.

Quote:

Originally Posted by FalconPUNCH!
I don't understand how you got that equation. Can you explain it to me? Sorry if it's asking too much.

Okay, so you have the slope of the tangent line at the point where x = 1. What is the y value of this point?

$y = 1^2e^{-1} = \frac{1}{e}$

Thus the tangent line at x = 1 has a slope of $\frac{1}{e}$ and touches the curve at the point $\left ( 1, \frac{1}{e} \right )$. (There are a couple of 1/e's floating around. Make sure you keep track of them.)

So your line equation, $y = mx + b$ becomes
$y = \frac{1}{e} \cdot x + b$
at the point $\left ( 1, \frac{1}{e} \right )$.

Plug this x and value into the line equation and solve for b.

-Dan
• Oct 22nd 2007, 07:23 PM
FalconPUNCH!
Quote:

Originally Posted by topsquark
Okay, so you have the slope of the tangent line at the point where x = 1. What is the y value of this point?

$y = 1^2e^{-1} = \frac{1}{e}$

Thus the tangent line at x = 1 has a slope of $\frac{1}{e}$ and touches the curve at the point $\left ( 1, \frac{1}{e} \right )$. (There are a couple of 1/e's floating around. Make sure you keep track of them.)

So your line equation, $y = mx + b$ becomes
$y = \frac{1}{e} \cdot x + b$
at the point $\left ( 1, \frac{1}{e} \right )$.

Plug this x and value into the line equation and solve for b.

-Dan

I understand that but I don't know how he got the derivative using the chain rule. I used the chain rule and got the answer in my original post. I don't know what I did wrong.
• Oct 22nd 2007, 07:39 PM
topsquark
Quote:

Originally Posted by FalconPUNCH!
I understand that but I don't know how he got the derivative using the chain rule. I used the chain rule and got the answer in my original post. I don't know what I did wrong.

Quote:

Originally Posted by galactus
That's the product rule, not the chain rule.

$\frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}$

If x=1, then the slope at that point is $\frac{1}{e}$

Now, use your given points in y=mx+b to find the line equation.

The product rule says if we have a function $f(x)g(x)$ then the derivative of that function is
$f^{\prime}(x)g(x) + f(x) g^{\prime}(x)$

Here we have $x^2 e^{-x}$, so
$f(x) = x^2 \implies f^{\prime}(x) = 2x$
and
$g(x) = e^{-x} \implies g^{\prime}(x) = -e^{-x}$

So the derivative of $x^2 e^{-x}$ will be:
$(2x)(e^{-x}) + (x^2)(-e^{-x}) = (2x - x^2)e^{-x}$

-Dan
• Oct 22nd 2007, 07:58 PM
FalconPUNCH!
Quote:

Originally Posted by topsquark
The product rule says if we have a function $f(x)g(x)$ then the derivative of that function is
$f^{\prime}(x)g(x) + f(x) g^{\prime}(x)$

Here we have $x^2 e^{-x}$, so
$f(x) = x^2 \implies f^{\prime}(x) = 2x$
and
$g(x) = e^{-x} \implies g^{\prime}(x) = -e^{-x}$

So the derivative of $x^2 e^{-x}$ will be:
$(2x)(e^{-x}) + (x^2)(-e^{-x}) = (2x - x^2)e^{-x}$

-Dan

Thank you for your help. I'll talk to my teacher because I'm not really sure what the book is asking. The section is about the chain rule so I assumed that you had to use it.
• Oct 22nd 2007, 08:01 PM
topsquark
Quote:

Originally Posted by FalconPUNCH!
I understand that but I don't know how he got the derivative using the chain rule. I used the chain rule and got the answer in my original post. I don't know what I did wrong.

Quote:

Originally Posted by galactus
That's the product rule, not the chain rule.

$\frac{d}{dx}[x^{2}e^{-x}]=(2x-x^{2})e^{-x}$

If x=1, then the slope at that point is $\frac{1}{e}$

Now, use your given points in y=mx+b to find the line equation.

You sort of do when you take the derivative of the exponential, but that's only when you don't block it out fully, like I did.

-Dan

Quote:

Originally Posted by FalconPUNCH!
Thank you for your help. I'll talk to my teacher because I'm not really sure what the book is asking. The section is about the chain rule so I assumed that you had to use it.