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!

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exponential differentiation question (1 Viewer)

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