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differentiate g(x) if $g(x)=e^xf(e^{-x})$

Using any website to evaluate this derivative like wolframalpha.com

we will get the result ===> $e^xf(e^{-x})-f'(e^{-x})$

But we know from the Product Rule of derivatives that it will like :

$e^xf'(e^{-x})+f(e^{-x})e^x$

why is there minus sign? and where the term $e^x$ has disappeared?

Very confusing !

Maher
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    Chain rule: the derivative of $f(e^{-x})$ is $- e^{-x}f'(e^{-x})$. Now put that back in the expression using the product rule and you'll recover the Wolfram answer. – Simon S Nov 20 '14 at 20:48
  • I know the chain rule .. but how is it applied to this kind functions of $f(e^{-x})$ .. I am not familiar with it..? – Maher Nov 20 '14 at 21:02

5 Answers5

5

You need to apply the chain rule when evaluating $\frac{d}{dx}f(e^{-x})$. Specifically, this derivative is $-e^{-x}f'(e^{-x})$. This extra factor of $e^{-x}$ cancels out the factor of $e^x$ and the negative sign materializes.

Reinstate Monica
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  • I know the chain rule .. but how is it applied to this kind functions of $f(e^{-x})$ .. I am not familiar with it – Maher Nov 20 '14 at 20:56
  • Write $h(x) = e^{-x}$. Then $f(e^{-x}) = f(h(x))$ so by the chain rule, $\frac{d}{dx} f(e^{-x}) = h'(x) f'(h(x)) = -e^{-x} f'(e^{-x})$. (Note that the derivative of $e^{-x}$ used the chain rule as well, which explains where the $-$ sign came from.) When taking the derivative of a nested expression, you should consider each layer as a function and use the chain rule. – Reinstate Monica Nov 20 '14 at 21:06
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$$\frac d{dx}\Big( f(e^{-x})\Big) = f'(e^{-x})\cdot \frac{d}{dx}(e^{-x}) = -1\cdot e^{-x} f'(e^{-x}) =- \dfrac {1}{e^x}f'(e^{-x})$$

That gives, for the second term $$(-1)e^x\cdot \frac 1{e^{x}}f'(e^{-x}) = -f'(e^{-x})$$

amWhy
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we have $g(x)=e^xf(e^{-x})$ then we get the first derivative by the product and the chain rule: $g'(x)=e^{x}f(e^{-x})+e^{x}f'(e^{-x})e^{-x}(-1)$ and since $e^{x}e^{-x}=1$ we can simplify the derivative $g'(x)=e^{x}f(e^{-x})-f'(e^{-x})$

Dr. Sonnhard Graubner
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$$g(x)=e^xf(e^{-x}) \implies g'(x) = e^x \left(-e^{-x}f'(e^{-x})\right) + e^x f(e^{-x})$$ This is just the product rule for $g = g_1g_2$, where $g_1(x) = e^x$ and $g_2(x) = f(e^{-x})$.

Ivo Terek
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I almost never use Lagrange's prime notation for derivatives because it doesn't explicitly show what is given respect to. For this reason, I recommend Leibniz's notation. So now $$ \frac{d}{dx}[g(x)]=\frac{d}{dx}\left[e^xf(e^{-x})\right] $$ $$= f(e^{-x})\frac{d}{dx}\left[e^x\right]+ e^{x}\frac{d}{dx}\left[f(e^{-x})\right] $$ $$= f(e^{-x})e^x+ e^{x}\frac{d}{d(e^{-x})}\left[f(e^{-x})\right]\frac{d}{dx}[e^{-x}]$$ $$= f(e^{-x})e^x+ e^{x}\frac{d}{d(e^{-x})}\left[f(e^{-x})\right]e^{-x}\frac{d}{dx}[-x]$$ $$= f(e^{-x})e^x+ e^{x}\frac{d}{d(e^{-x})}\left[f(e^{-x})\right]e^{-x}(-1)$$ $$= f(e^{-x})e^x- e^{x}\frac{d}{d(e^{-x})}\left[f(e^{-x})\right]e^{-x}$$ $$= f(e^{-x})e^x- \frac{d}{d(e^{-x})}\left[f(e^{-x})\right]$$

k170
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