Question

Tentukan turunan dari
[tex]\rm f(x)=\sqrt[4]{(x^3-3x^2+4)^3}[/tex]​

Answer 1

Jawaban:

turunan fungsi aljabar

[tex]f(x) = {ax}^{p} \to \: f'(x) = p.ax {}^{p - 1} \\ [/tex]

[tex]f(x) = \sqrt[4]{ ({x}^{3} - 3 {x}^{2} + 4) {}^{3} } \\ f(x) = ( {x}^{3} - 3 {x}^{2} + 4) {}^{ \frac{3}{4} } \\ f'(x) = \frac{3}{4} ( {x}^{3} - 3 {x}^{2} + 4) {}^{ \frac{3}{4} - 1 } .(3 {x}^{2} - 6x) \\ f'(x) = \frac{3}{4} (3 {x}^{2} - 6x)( {x}^{3} - 3 {x}^{2} + 4) {}^{ - \frac{1}{4} } \\ f'(x) = \frac{3(3 {x}^{2} - 6x)}{4( {x}^{3} - 3 {x}^{2} + 4) {}^{ \frac{1}{4} } } \\ f'(x) = \frac{9 {x}^{2} - 18x }{4( {x}^{3} - 3 {x}^{2} + 4) {}^{ \frac{1}{4} } } \\ f'(x) = \frac{9 {x}^{2} - 18x}{4 \: \sqrt[4]{ {x}^{3} - 3 {x}^{2} + 4} } [/tex]

Answer 2

Jawab:

[tex]\displaystyle \frac{9\sqrt{x-2}}{4}~\sqrt[4]{\frac{x^4}{x+1}}[/tex]

Penjelasan dengan langkah-langkah:

Aturan rantai [tex]\displaystyle \frac{dy}{dx}=\frac{dy}{du}~\frac{du}{dx}[/tex]. Berdasarkan bentuk akar [tex]\displaystyle \sqrt[n]{a^m}=a^{\frac{m}{n}}[/tex] maka [tex]\displaystyle y=(x^3-3x^2+4)^{\frac{3}{4}}[/tex]

[tex]\displaystyle \begin{matrix}y=u^{\frac{3}{4}} & u=x^3-3x^2+4\\ \frac{dy}{du}=\frac{3}{4}u^{-\frac{1}{4}} & \frac{du}{dx}=3x^2-6x\end{matrix}[/tex]

Diperoleh

[tex]\begin{aligned}\frac{dy}{dx}&\:=\frac{dy}{du}~\frac{du}{dx}\\\:&=\frac{3}{4}u^{-\frac{1}{4}}(3x^2-6x)\\\:&=\frac{9x^2-18x}{\sqrt[4]{u}}\\\:&=\frac{9(x^2-2x)}{4\sqrt[4]{x^3-3x^2+4}}\\\:&=\frac{9x(x-2)}{4\sqrt[4]{x^3-4x^2+x^2-4x+4x+4}}\\\:&=\frac{9x(x-2)}{4\sqrt[4]{(x^2-4x+4)(x+1)}}\\\:&=\frac{9x(x-2)}{4\sqrt[4]{(x-2)^2(x+1)}}\\\:&=\frac{9x(x-2)}{4\sqrt[4]{x+1}~\sqrt{x-2}}\\\:&=\frac{9\sqrt{x-2}}{4}~\sqrt[4]{\frac{x^4}{x+1}}\\\end{aligned}[/tex]