Question

Find an equation of the tangent to the curve y = e^x that passes thru the origin.

Answer 1

So first we use the derivative to find the slope. Can you do that?

Answer 2

Now we know \[\Large m=\frac{y_2-y_1}{x_2-x_1}\]
So \[\Large \frac{dy}{dx}e^x=\frac{e^x-0}{x- 0}\]

Answer 3

is that the answer?

Answer 4

No, but using this equation we can solve for \(x\), to find which point to use to get the slope.

Answer 5

Then we can put the line in the form \( y = mx +b \) and we know \(b=0\) since the origin is the intercept.

Answer 6

so the answer is e^x

Answer 7

No, \(\large e^x\) isn't even a line! You need to solve for \(x\) in \[\Large \frac{dy}{dx}e^x=\frac{e^x-0}{x- 0}\]

Answer 8

First start of by finding the derivative.

Answer 9

the derivative is e^x

Answer 10

and then e^0 is 1

Answer 11

Okay so when does \[\large e^x = \frac{e^x}{x}\]
What must x be for both sides to be equal?

Answer 12

1

Answer 13

Good, so now we can use it to find \(m\) by plugging \(1\) into \(\large e^x\).

Answer 14

And our answer will be \(y = mx+b\). We know \(b = 0\) and \(m = e^1\).

Answer 15

So the line is \(y=e \cdot x\). Make sense?

Answer 16

yes...Thank You

Answer 17

http://www.wolframalpha.com/input/?i=e%5Ex%2Cex

And this confirms it!

Answer 18

Thank you ....Have a nice day :)