Proof of $f(x)=\sinh x \Rightarrow \frac{df}{dx}=\cosh x$ with definition of derivative.

Question

I'm trying to prove that,

$$f(x)=\sinh x=\frac{e^{x}-e^{-x}}{2} \implies \frac{df}{dx}=\cosh x=\frac{e^{x}+e^{-x}}{2}$$

using the definition of derivative.

\begin{align}f(x)=\sinh x =\frac{e^{x}-e^{-x}}{2} &\frac{df}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ &=\lim_{h\to 0}\frac{\frac{e^{x+h}-e^{-(x+h)}}{2}-\frac{e^{x}-e^{-x}}{2}}{h} \\ &=\lim_{h\to 0}\frac{e^{x+h}-e^{-x-h}-e^{x}+e^{-x}}{2h} \\ &=\lim_{h\to 0}\frac{e^{x}e^{h}-e^{-x}e^{-h}-e^{x}+e^{-x}}{2h} \\ &=\lim_{h\to 0}\frac{e^{x}(e^{h}-1)+e^{-x}(-e^{-h}+1)}{2h} \\ \end{align}

And I can't go anymore from here. How can I prove it?

score 2 · Accepted Answer · answered Nov 23 '19 at 13:43

2

Now use the fact that$$\lim_{h\to0}\frac{e^h-1}h=1\text{ and }\lim_{h\to0}\frac{e^{-h}-1}h=-1,$$both of which follow from the fact that $\exp'(0)=1$.

answered Nov 23 '19 at 13:43

José Carlos Santos

427,504

Oh, I remember that $\lim_{h\to0}\frac{e^{h}-1}{h}=1$ and $\lim_{h\to0}\frac{e^{-h}-1}{h}=-1$. But I wonder how it is equal to $1$ or $-1$. Can you help me about it? – M. Çağlar TUFAN Nov 23 '19 at 13:48
That is another question. I suggest that you post it as such. – José Carlos Santos Nov 23 '19 at 13:54
@JoséCarlosSantos It's an easy duplicate. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Nov 23 '19 at 14:02
@TheSimpliFire Sure. I should have thought that. – José Carlos Santos Nov 23 '19 at 14:05
@JoséCarlosSantos Thank you for your help. I'm happy with your answer :) – M. Çağlar TUFAN Nov 23 '19 at 14:11

Proof of $f(x)=\sinh x \Rightarrow \frac{df}{dx}=\cosh x$ with definition of derivative.

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