# Thread: Equation for Tangnet line, Logs

1. ## Equation for Tangnet line, Logs

Hi, I am having trouble with this problem. Directions are: Find an equation for the tangent line to y=f(x) at the specified point.

f(x)= e^2x / x^2 ; Where x= 1

I used the quotient rule and obtained: ((x^2)(2e^2x) - (e^2x)(2x)) / x^4
I hope I am correct up to here, I don't know what to do from here to get the equation for the tangent line, the book has the answer as: y=e^2 If someone could illustrate how this is obtained, Thank You.

2. Hello,

A tangent line to a function has a direction coefficient equal to the derivate number at this point.

At point of abscissa 1, the equation of the tangent line will be :

$y=f'(1)(x-1)+f(1)$

Your derivative is correct (you can factorise by $2xe^{2x}$ if you want to make it more beautiful ), so find its value at the point of abscissa 1.

3. Could you please show how I get the final answer? I think I have to use the point-slope formula, but aren't certain. Thanks again.

4. You just have to use the formula i gave you...

Then, just calculate f '(1) and f(1) to have the complete formula ! Is it really useful to give you all the answer ? You're close to it !

6. Ok...

So you got the formula of the derivative :

$f'(x)=\frac{(x^2)(2e^{2x}) - (e^{2x})(2x)}{x^4} = \dots = \frac{2x^2 e^{2x}-2xe^{2x}}{x^4}$

Hence $f'(1)=\frac{2e^2 - 2e^2}{1}=0$

-> the equation of the tangent line at point of abscissa 1 is :

$y=f'(1)(x-1)+f(1)=0*(x-1)+f(1)=f(1)$

$f(x)=\frac{e^{2x}}{x^2}$

Hence $f(1)=\frac{e^{2*1}}{1^2}=e^2$

-> $y=e^2$ at the point we're studying.

7. Couldn't I have just plugged in a 1 for the x variables in the original equation? What does the formula for the tangent line tell us? The derivative goes to zero, so I assume this means the points are (1,0), so did you plug these into the formula? Then what happens after that? I am probably making this harder than it really is, I just can't "see" how it all comes together.

8. Hmm, i'll try to explain.

If you replace x with 1 in the expression of f(x) then derivate, you will got 0, because f(1) is a number.

You have to derivate first f(x), then replace x by 1 in f '(1).

The point (1,0) is not on the initial curve (representing f). Graphically, f '(1)=0 only means that the slope of the tangent line is 0, nothing about the point (1,0)

I hope it's a little bit clear :S

The only formula to use to find the equation of the tangent line at the point of abscissa a is y=f '(a)(x-a)+f(a)