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[tex]\displaystyle\sf\frac{{d}^{2}}{{dx}^{2}}(\int\limits_{1}^{x}8x{({x}^{2}+1)}^{3}\:dx-\lim\limits_{h\to0}\frac{8{(x+h)}^{4}+{8x}^{4}}{h})=...[/tex]​

Berikut ini adalah pertanyaan dari unknown pada mata pelajaran Matematika untuk jenjang Sekolah Dasar

\displaystyle\sf\frac{{d}^{2}}{{dx}^{2}}(\int\limits_{1}^{x}8x{({x}^{2}+1)}^{3}\:dx-\lim\limits_{h\to0}\frac{8{(x+h)}^{4}+{8x}^{4}}{h})=...​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

\begin{aligned}&\frac{d^2}{dx^2}\left(\int_1^x{8x\left(x^2+1\right)^3dx}\:-\:\lim_{h\to0}\frac{8{(x+h)}^{4}-{8x}^{4}}{h}\right)\\&=\ \large\text{$\bf56x^6+120x^4+72x^2-192x+8$}\end{aligned}

Pembahasan

(Karena sudah dikoreksi pada kolom komentar, sehingga soal ini jelas arahnya, maka saya coba kerjakan.)

\begin{aligned}&\frac{d^2}{dx^2}\left(\int_1^x{8x\left(x^2+1\right)^3dx}\:-\:\lim_{h\to0}\frac{8{(x+h)}^{4}-{8x}^{4}}{h}\right)\\\\&{=\ }\frac{d^2}{dx^2}\left(\int_1^x{8x\left(x^2+1\right)^3dx}\right)\:-\:\frac{d^2}{dx^2}\underbrace{\left(\lim_{h\to0}\frac{8{(x+h)}^{4}-{8x}^{4}}{h}\right)}_{\begin{array}{c}{\sf D{ef}inisi}\ \frac{d}{dx}\left(8x^4\right)\end{array}}\end{aligned}

\begin{aligned}&{=\ }\frac{d}{dx}\left(\frac{d}{dx}\int_1^x{8x\left(x^2+1\right)^3dx}\right)\:-\:\frac{d^3}{dx^3}\left(8x^4\right)\\\\&{=\ }8\left[\:\frac{d}{dx}\left(\frac{d}{dx}\int_1^x{x\left(x^2+1\right)^3dx}\right)\:-\:\frac{d^3}{dx^3}\left(x^4\right)\:\right]\end{aligned}

\begin{aligned}&{.\quad}\left[\ \begin{aligned}&\textsf{Jika $g(x)=f'(x)$, maka}\\&\quad\frac{d}{dx}\int_c^x{g(x)\,dx}\ \textsf{($c$ adalah konstanta)}\\&\quad=\frac{d}{dx}\Bigl[f(x)-f(c)\Bigr]\\&\quad=\frac{df(x)}{dx}-\frac{df(c)}{dx}\\&\quad=\frac{df(x)}{dx}-0\\&\quad=\frac{df(x)}{dx}\\&\quad=g(x)\end{aligned}\right.\end{aligned}

\begin{aligned}&{=\ }8\left[\:\frac{d}{dx}\left(x\left(x^2+1\right)^3\right)\:-\:\frac{d^3}{dx^3}\left(x^4\right)\:\right]\\\\&{=\ }8\left[\:(x)'\left(x^2+1\right)^3\:+\:x\left(\left(x^2+1\right)^3\right)'\:-\:4\cdot3\cdot2\cdot x^{(4-3)}\:\right]\\\\&{=\ }8\left[\:\left(x^2+1\right)^3\:+\:x\cdot3\left(x^2+1\right)^2\left(x^2+1\right)'\:-\:24x\:\right]\\\\&{=\ }8\left[\:\left(x^2+1\right)^3\:+\:3x\left(x^2+1\right)^2(2x)\:-\:24x\:\right]\end{aligned}

\begin{aligned}&{=\ }8\left[\:\left(x^2+1\right)^3\:+\:6x^2\left(x^2+1\right)^2\:-\:24x\:\right]\\\\&{=\ }8\left[\:\left(x^2+1+6x^2\right)\left(x^2+1\right)^2\:-\:24x\:\right]\\\\&{=\ }8\left[\:\left(7x^2+1\right)\left(x^4+2x^2+1\right)\:-\:24x\:\right]\\\\&{=\ }8\left[\:7x^6+14x^4+7x^2+x^4+2x^2+1\:-\:24x\:\right]\\\\&{=\ }8\left[\:7x^6+15x^4+9x^2+1\:-\:24x\:\right]\\\\&{=\ }\boxed{\ \bf56x^6+120x^4+72x^2-192x+8\ }\end{aligned}

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Last Update: Sat, 02 Jul 22
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