# Tangent line to curve, derivative.

## Homework Statement

See the attachment. I feel like my answer is right, but I keep getting told I am wrong. This is the first time I have taken any math in 5 years so I am figuring I must not be simplifying enough??? But I can't see what I need to simplify. Perhaps I have made a dumb error and I'm just not seeing it. ARR. I have put this response in a million different ways but I keep getting told I'm wrong, please help! I've tried taking the e^2x out. Ive tried multiplying the x through the parenthisis, tried brackets instead of parenthesis. What am i missing?

## The Attempt at a Solution

#### Attachments

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I'm not sure why you attached that x+1 to the end but the expression that's attached is just the derivativ of the whole function. You're looking for the slope at a single point only, that is (0,1), and then you can go ahead and find an equation of a tangent line using the slope and a point.

So you're next step should be to evaluate the derivative at the point you need the slope for.

Hope that helps.

hmmm perhaps I should so my work....

I got the derivative, so m, to equal (e^2x)(-pisin(pix)) + (2e^2x)(cos(pix))

and then using the point slope formula

y-1=m(x-0)....

and then i plugged in what I got for the derivative for m

which gave me the answer I had posted above, and moving the -1 to the other side which makes it +1.

Am I totally off base?

Char. Limit
Gold Member
hmmm perhaps I should so my work....

I got the derivative, so m, to equal (e^2x)(-pisin(pix)) + (2e^2x)(cos(pix))

and then using the point slope formula

y-1=m(x-0)....

and then i plugged in what I got for the derivative for m

which gave me the answer I had posted above, and moving the -1 to the other side which makes it +1.

Am I totally off base?
You don't plug in the derivative. You plug in the value of derivative at the point specified.

OMG THANK U! i was making my life so much harder than it needed to be! :)