Calcul of tangent line

[Carlito]

Hi. i having a little trouve to resolve those 2 equations.
i wont lie i am almost at point 0 for those 2.

Question 5 and 6.

For question 5 i understand that i have to take the derivative of the equation and replace x by the x they give on the point and so on .

My problem is to find the derivative of that equation .
i get : e^t++1

any help would be appreciacce i am block on those 2.

 

[daon2]

Your posted picture says it's an exam? Would help received for these problems be considered cheating?

I will tell you that you are having trouble applying the chain rule, that would be the place to start.

 

[Carlito]

thank you for your answer, these are not exam i will have to do. They are old midterm from 2012 available at the printer store , for pratice they are not from this year.

 

[Carlito]

why do we need to use the chain rules ? shoulnt it be a D/Dx of every term then adding them its a sum of derivative .... sorry i am kind of lost

 

[daon2]

why do we need to use the chain rules ? shoulnt it be a D/Dx of every term then adding them its a sum of derivative .... sorry i am kind of lost

Let's start with the first term, \(\displaystyle f(t)=e^t\). Since \(\displaystyle t\) is a function of \(\displaystyle x\), we need to think of \(\displaystyle t\) as the function \(\displaystyle t(x)\). When we apply \(\displaystyle \dfrac{d}{dx}\) we get:

\(\displaystyle \dfrac{d}{dx} f(t) = f'(t)\cdot \dfrac{d}{dx}(t(x)) = e^t\cdot t'(x)\).

When you do this to every term, you can then solve for \(\displaystyle t'(x)\) and plug in \(\displaystyle x=e, t=1\).

 

[Carlito]

thx men. i should have taught directly to use implicit differentiation will pratice that more.

the way we notes the stuff are a bit different from you.

Now that i have the derivative.

the formula of the slope is M= y2-y1/x2-x1
But i only have one points.
Do i replace T by 0 on the t" to find an another point ?

Forget that i just get it ... you the best men thx

 

[daon2]

thx men. i should have taught directly to use implicit differentiation will pratice that more.

the way we notes the stuff are a bit different from you.

Now that i have the derivative.

the formula of the slope is M= y2-y1/x2-x1
But i only have one points.
Do i replace T by 0 on the t" to find an another point ?

Forget that i just get it ... you the best men thx

To find the slope, you use the derivative you solved for. You have a point \(\displaystyle (x,t) = (e,1)\). Then \(\displaystyle t'(e)\) is the slope of the tangent line at this point