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I have to prove the derivative by definition of

$$\frac{\sin^2(x)}{e^x-1}$$

$$f^{\prime}(x)=\lim_{\Delta x \to 0}{\frac{f(x+\Delta x)-f(x)}{\Delta x}}$$

$$\large f^{\prime}(x)=\lim_{\Delta x \to 0}{\frac{\frac{\sin^2(x+\Delta x)}{e^{(x+\Delta x)}-1}-\frac{\sin^2(x)}{e^x-1}}{\Delta x}}$$

I tried to rid of ${e^x-1}$ in different ways but with no luck. Actually the problem is the limit of the last expression.

Teoc
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Antonio Velazquez Bustamante
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1 Answers1

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we let $\Delta x=t$,then you only find this limits $$\lim_{t\to 0}\dfrac{\sin^2{(x+t)}e^x-\sin^2{(x+t)}-\sin^2{x}e^{x+t}+\sin^2{x}}{t}$$ and \begin{align*}&\sin^2{(x+t)}e^x-\sin^2{x}e^x+\sin^2{x}e^x-\sin^2{x}e^{x+t}-[\sin^{(x+t)}-\sin^2{x}]\\ &=(e^x-1)[\sin^2{(x+t)}-\sin^2{x}]+\sin^2{x}e^x[1-e^t]\\ &=(e^x-1)\sin{(2x+t)}\sin{t}-\sin^2{x}e^x[1-e^t]\\ &=((e^x-1)\sin{2x}-\sin^2{x}e^x)\cdot t+o(t) \end{align*} so $$\lim_{\Delta \to 0}{\frac{\frac{\sin^2(x+\Delta x)}{e^{(x+\Delta x)}-1}-\frac{\sin^2(x)}{e^x-1}}{\Delta x}}=\lim_{t\to 0}\dfrac{\dfrac{\sin^2{(x+t)}(e^x-1)-\sin^2{x}(e^{x+t}-1)}{(e^{x+t}-1)(e^x-1)}}{t}=\dfrac{(e^x-1)\sin{2x}-\sin^2{x}e^x}{(e^x-1)^2}$$

where we use following indenty $$\sin^2{x}-\sin^2{y}=\sin{(x+y)}\sin{(x-y)}$$ and $$\lim_{t\to 0}\dfrac{\sin{t}}{t}=1,\lim_{t\to 0}\dfrac{e^t-1}{t}=1$$ you can solve it

math110
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