Nanya dua pertanyaan ya ^_^.... Question from @nokitron

October 2018 1 119 Report

Nanya dua pertanyaan ya ^_^

Takamori37 Soal kiri.
Ubah menjadi bentuk e.
$$\begin{align}y&=x^{ax^2} \\ y&=(e^{\ln x})^{ax^2} \\ y&=e^{ax^2.\ln x} \\ &\text{Sehingga:} \\ \frac{dy}{dx}&=e^{ax^2.\ln x}(ax^2.\ln x)' \\ &=e^{ax^2.\ln x}\left[2ax\ln x+ax^2.\frac1x\right] \\ &=e^{ax^2.\ln x}(2ax\ln x+ax) \\ &=ax.e^{ax^2.\ln x}(2\ln x+1)\end{align}$

Soal kanan.
$$\begin{align}&\lim_{x\to0}\csc^2(2x)-\frac1{4x^2} \\ &=\lim_{x\to0}\frac{4x^2\csc^2(2x)-1}{4x^2} \\ &=\lim_{x\to0}\frac{8x\csc^2(2x)+4x^2.2\csc(2x).(-2\csc(2x)\cot(2x))}{8x}&\boxed{?} \\ &=\lim_{x\to0}\frac{8x\csc^2(2x)-16x^2\csc^2(2x)\cot(2x)}{8x} \\ &=\lim_{x\to0}\csc^2(2x)-2x\csc^2(2x)\cot(2x) \\ &=\lim_{x\to0}\csc^2(2x)(1-2x\cot(2x)) \\ &=\lim_{x\to0}\frac{1-2x.\cot(2x)}{\sin^2(2x)}\end{align}$
Di L'Hopital beberapa kali.
$$\begin{align}&=\lim_{x\to0}\frac{-(2\cot(2x)+2x.(-2\csc^2(2x))}{4\sin(2x)\cos(2x)} \\ &=\lim_{x\to0}\frac{4x\csc^2(2x)-2\cot(2x)}{4\sin(2x)\cos(2x)} \\ &=\lim_{x\to0}\frac{4x/\sin^2(2x)-2\cos(2x)/\sin(2x)}{2\sin(4x)} \\ &=\lim_{x\to0}\frac{4x-2\sin(2x)\cos(2x)}{2\sin^2(2x)\sin(4x)} \\ &=\lim_{x\to0}\frac{4x-\sin(4x)}{2\sin^2(2x)\sin(4x)}\end{align}$
Lanjut.
$$\begin{align}&=\lim_{x\to0}\frac{4x-\sin(4x)}{[(2\sin^2(2x)-1)+1]\sin(4x)} \\ &=\lim_{x\to0}\frac{4x-\sin(4x)}{(1-\cos(4x))\sin(4x)} \\ &=\lim_{x\to0}\frac{[4x-\sin(4x)][1+\cos(4x)]}{[1-\cos^2(4x)]\sin(4x)} \\ &=\lim_{x\to0}\frac{(4x-\sin(4x))(1+\cos(4x))}{\sin^3(4x)} \\ &=\lim_{x\to0}\frac{2(4x-\sin(4x))}{\sin^3(4x)} \\ &=\lim_{x\to0}\frac{2(4-4\cos(4x))}{12\sin^2(4x)\cos(4x)} \\ \end{align}$
Lanjutkan lagi.
$$\begin{align}&=\lim_{x\to0}\frac{8(1-\cos(4x))}{12\sin^2(4x)\cos(4x)} \\ &=\lim_{x\to0}\frac{2(2\sin^2(2x))}{3\sin^2(4x)} \\ &=\lim_{x\to0}\frac{4}{3}\left[\frac{\sin(2x)}{\sin(4x)}\right]^2 \\ &=\frac43\times\frac{1}{2^2} \\ &=\frac13\end{align}$

1 votes Thanks 2

nokitron Terima kasih Mau nanya lagi, kak. Kalo jawaban no. 1 bisa disederhanakan lagi tidak?

Takamori37 Boleh, balik lagi. ax. x^(ax^2).(2ln x + 1)

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Bentuk sederhana dari √(33+√800) - √(27-2√162) adalah

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