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Use the limit definition of derivative to find $\frac{\mathrm d}{\mathrm dx}f(x)$ for the function $f(x)= e^{2(x+1)}$.

I know it's going to be $2e^{2x+2}$. I can solve it normally, I just don't know how to solve it using the definition, please help.

k170
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Tobias
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2 Answers2

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Following the definition we get : \begin{align*} f'(x) ={}& \lim_{h\rightarrow 0} {f(x+h)-f(x)\over h} \\ {}={}& \lim_{h\rightarrow 0} {e^{2(x+h+1)}-e^{2(x+1)}\over h} \\ {}={}& \lim_{h\rightarrow 0} {e^{2h}e^{2(x+1)}-e^{2(x+1)}\over h} \\ {}={}& e^{2(x+1)} \lim_{h\rightarrow 0} {e^{2h}-1\over h} \\ {}={}& e^{2(x+1)} \lim_{h\rightarrow 0} {1 \over h} \left(2h+{(2h)^2 \over 2!} +{(2h)^3 \over 3!} + \cdots \right) \\ {}={}& 2 \cdot e^{2(x+1)} \\ \end{align*}

MickG
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nickchalkida
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    Alternatively, instead of going to the power series, one could multiply and divide by $2$ with a change of variables where $k=2h$. Many students see/know $\lim_{k\rightarrow 0}\frac{e^k-1}{k}=1$ before power series. – Michael Burr Feb 17 '16 at 14:21
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$$\frac{\mathrm d}{\mathrm dx}\left[e^{2(x+1)}\right] $$ $$= \lim\limits_{h\to 0} \frac{e^{2(x+h+1)}-e^{2(x+1)}}{h}$$ $$= \lim\limits_{h\to 0} \frac{\left(e^{2}\right)^{(x+h+1)}-\left(e^{2}\right)^{(x+1)}}{h}$$ $$= \lim\limits_{h\to 0} \frac{\left(e^{2}\right)^{(x+1)}\left(e^{2}\right)^h-\left(e^{2}\right)^{(x+1)}}{h}$$ $$= \left(e^{2}\right)^{(x+1)}\lim\limits_{h\to 0} \frac{\left(e^{2}\right)^h-1}{h}$$ $$= 2e^{2(x+1)}\lim\limits_{h\to 0} \frac{e^{2h}-1}{2h}$$ Let $t=2h$, so now we have $$2e^{2(x+1)}\lim\limits_{t\to 0} \frac{e^{t}-1}{t}$$ $$=2e^{2(x+1)}\ln e$$ $$=2e^{2(x+1)}$$

k170
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