# Am I doing this right?

• Oct 27th 2009, 05:57 PM
skboss
Am I doing this right?
Let f(x)= xe^(x). Use the limit definition to compute f'(0) and find the equation of the tangent line at x=0

f(x)= xe^(x)

limit definition f'(x)= f(x+h)-f(x) / h as lim h-->0

so

f'(x) = (x+h)* (e^(x+h) - xe^(x) / h

f'(0) = (0+h)* (e^(0+h) - xe^(0) / h

= h*e(h)- 0 / h

= e^(h) = e(0) =1

So, the derivative is 1.

Is this right?
• Oct 27th 2009, 06:06 PM
skeeter
Quote:

Originally Posted by skboss
Let f(x)= xe^(x). Use the limit definition to compute f'(0) and find the equation of the tangent line at x=0

f(x)= xe^(x)

limit definition f'(x)= f(x+h)-f(x) / h as lim h-->0

so

f'(x) = (x+h)* (e^(x+h) - xe^(x) / h

f'(0) = (0+h)* (e^(0+h) - xe^(0) / h

= h*e(h)- 0 / h

= e^(h) = e(0) =1

So, the derivative is 1.

Is this right?

close enough from what I can discern from your syntax ... this might be a bit easier.

$\displaystyle f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$

$\displaystyle f'(0) = \lim_{x \to 0} \frac{xe^x - 0}{x - 0}$

$\displaystyle f'(0) = \lim_{x \to 0} \frac{xe^x}{x}$

$\displaystyle f'(0) = \lim_{x \to 0} e^x = e^0 = 1$
• Oct 27th 2009, 06:27 PM
skboss
Thanks. BTW, Can I use that a limit definition to find derivative? I wasn't sure about it.