exponential differentiation question (1 Viewer)

SunnyScience · Feb 4, 2012

If f(x) = a/2(e^(x/a) + e^(-x/a))
Find f'(a).

Thanks

Carrotsticks · Feb 4, 2012

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

Is this the function?

Or is it this?

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

IamBread · Feb 4, 2012

SunnyScience said:
If f(x) = a/2(e^(x/a) + e^(-x/a))
Find f'(a).

Thanks

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

$$Using $ \frac{d}{dx}e^{g(x)} = g'(x)e^{g(x)}$

$f'(x) = \frac{a}{2}(\frac{1}{a}e^{\frac{x}{a}} + \frac{-1}{a}e^{\frac{-x}{a}})$

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

$f'(a) = \frac{1}{2}(e - \frac{1}{e})$

SunnyScience · Feb 4, 2012

IamBread said:
$f(x) = \frac{a}{2}(e^{\frac{x}{a}} + e^{\frac{-x}{a}})$

$$Using $ \frac{d}{dx}e^{g(x)} = g'(x)e^{g(x)}$

$f'(x) = \frac{a}{2}(\frac{1}{a}e^{\frac{x}{a}} + \frac{-1}{a}e^{\frac{-x}{a}})$

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

$f'(a) = \frac{1}{2}(e - \frac{1}{e})$

Can you please explain how you got from the third last step to the second last please :A just started this topic >->

Carrotsticks · Feb 4, 2012

Oh dear, it was neither of what I typed out haha. I missed out a set of brackets.

IamBread · Feb 4, 2012

SunnyScience said:
Can you please explain how you got from the third last step to the second last please :A just started this topic >->

You just sub a's into the x's. Say, for example,
$f(x) = x^2 + x + 3, $ Then, $ f(a) = a^2 + a + 3, f(b) = b^2 + b + 3, f(z) = z^2 + z + 3 $ etc.$$

SunnyScience · Feb 4, 2012

Thanks anyway sticks

SunnyScience · Feb 4, 2012

IamBread said:
You just sub a's into the x's. Say, for example,
$f(x) = x^2 + x + 3, $ Then, $ f(a) = a^2 + a + 3, f(b) = b^2 + b + 3, f(z) = z^2 + z + 3 $ etc.$$

BUT OF COURSE! I must be getting tired sorry

thanks for your solution!

exponential differentiation question (1 Viewer)

SunnyScience

Member

Carrotsticks

Retired

IamBread

Member

SunnyScience

Member

Carrotsticks

Retired

IamBread

Member

SunnyScience

Member

SunnyScience

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)