dy/dx

T

Tueseve728

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Sep 3, 2005
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are there any points (ordered pairs) on the curve y=x-e^-x where the slope is 2? If so, find them.
This is how far i got,
y=x-e^-x
dy/dx=1+e-x
1+e^-x=2
e^-x=1
lne^-x=ln 1
-x=ln 1
x=0
now how do i determine the points?
 
What was the purpose of finding the derivative?

What does x = 0 mean for this problem?
 
a) What is the relationship between the derivative and "slope at a point"?

b) When you set dy/dx equal to "2", what were you trying to find?

c) What is the meaning, in the context of the above, of your solution "x = 0"?

Eliz.
 
Yes, you set dy/dx = 2 because you wanted to find when the slope was 2.

You did the problem correctly, but the final step is to find the point since all you know right now is the x value.

Luckily you know what the function is:
y=x-e^-x

Just plug in your x value and you will have a point where the slope is 2


Ted
 
ok thanks so much for ur help, I have one more questions...
find dy/dx of y=ln(2^tan x)
f(x)=ln(x)
g(x)=2^x
h(x)=tan(x)
then...
f prime(1/x)
g prime (2^x) (ln 2)
h prime (sec x^2)

then i get y prime =(1/(2^tanx))(2^tan x)(ln 2)(sec x^2)

this is right?
 
Please post new questions as new threads, not as replies to old threads.

Please review the replies you received to your other two posts of this exercise, here and here.

Please post follow-ups to one of those threads. Thank you.

Eliz.
 
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