find equation of the tangent to the graph

[Ashley5]

Find the equation of the tangent to the graph Please help, Thank you!


f(x)=e^x x=2

I know you first have to take the derivative
f(x+h)=e^(x+h)
f(x+h)-f(x)=e^(x+h)-(e^x) this is the step I'm having trouble with
f(x+h)-f(x)/h
f'(x)=lim f(x+h)-f(x)/h
h->0

to get the slope sub f'(2)=?

Then to get the equation in slope intercept form: y-y1=m(x-x1)


I know the answer is y=e^2 (x) -e^2

 

[tkhunny]

You have to start with this:

\(\displaystyle \frac{e^{x+h}=e^{x}}{h}\;=\;\frac{e^{x}e^{h}-e^{x}}{h}\;=\;e^{x}\frac{e^{h}-1}{h}\)

The only trick is not that list, simpler-looking limit. That is not such an easy task.

Here's an interesting way, but you may not like it:

http://mathforum.org/library/drmath/view/60705.html

 

[jwpaine]

Find tangent to graph f(x)=e^x at x=2

We know that f'(x) = e^x because of logarithmic differentiation:

f(x) = e^x
ln(f(x)) = ln(e^x)
ln(f(x)) = x
ln'(f(x)) = x'
[1/f(x)]f'(x) = 1
f'(x) = f(x)
f'(x) = e^x

so....

f'(x) = e^x
slope = f'(2) = e^2

now you have point (2, f(2))

Use point-slope: y - f(2) = f'(2)(x - 2)

Put it in slope-intercept form, then your done.

 

[Ashley5]

thank you! For your help