Aku & Maths: Takkan Terpisah (LEVEL 3)

y'= $$\frac{1}{\sqrt[]{3x^3}}$$

y'= $$\frac{-1}{\sqrt[]{\left(3x\right)^3}}$$

y'= $$\frac{1}{\sqrt[]{\left(3x\right)^3}}$$

y'= $$\frac{-1}{\sqrt[]{3x^3}}$$

 $3x^2-2e^x\left(x+1\right)$  

 $3x^2-2\left(e^x+\frac{x}{e^x}\right)$  

 $3x^2-2xe^x\left(2x+1\right)$  

 $3x^2-2x\left(2xe^x+\frac{1}{e^x}\right)$  

$$\sqrt[3]{-6}$$

$$\sqrt[3]{6}$$

$$\sqrt[3]{-3}$$

$$\sqrt[3]{3}$$

$$-\frac{3}{2\sqrt[]{x^3}}-\frac{5}{2\sqrt[]{x}}$$

$$-\frac{3}{2\sqrt[]{x^3}}+\frac{5}{2\sqrt[]{x}}$$

$$\frac{3}{2\sqrt[]{x^3}}-\frac{5}{2\sqrt[]{x}}$$

$$\frac{3}{2\sqrt[]{x^3}}+\frac{5}{2\sqrt[]{x}}$$

$$f'\left(x\right)=3+\frac{3}{2x^2}$$  

$$f'\left(x\right)=3-\frac{3}{2}x^{-2}$$  

$$f'\left(x\right)=\frac{12x+4}{2}$$  

$$f'\left(x\right)=3+\frac{2}{3x^2}$$