미적분 Unit 2 Test Review 2.1-3.1

$$f'(x)=18x^5-3x^2$$

$$f'(x)=18x^6-3x^2$$

$$f'(x)=18x^4-3x$$

$$f'(x)=6x^5-3x^3$$

$$f'(x)=\cos(x)\cdot\sin(\sin(x))$$

$$f'(x)=-\cos(\sin(x))\cdot\sin(x)$$

$$f'\left(x\right)=\cos\left(\sin\left(x\right)\right)\cdot\cos\left(x\right)$$

$$f'(x)=\sin(\sin(x))\cdot\sin(x)$$

$$f'(x)=e^x(2x-5)$$

$$f'(x)=e^x(2x-2)$$

$$f'(x)=e^x(2x-4)$$

$$f'(x)=e^x(2x-3)$$

$$\frac{2x\cos\left(x\right)-\sin\left(x\right)}{6x^{\frac{3}{2}}}$$

$$\frac{2x\cos\left(x\right)-\sin\left(x\right)}{9\sqrt{\ x}}$$

$$\frac{3\cos\left(x\right)\sqrt{\ x}}{9x}-\frac{3\sin\left(x\right)}{2\sqrt{\ x}}$$

$$3\cos\left(x\right)\sqrt{\ x}+\frac{3\sin\left(x\right)}{2\sqrt{\ x}}$$

$$\frac{2x\cos\left(x\right)-\sin\left(x\right)}{6x^{\frac{3}{2}}}$$

$$\frac{2x\cos\left(x\right)-\sin\left(x\right)}{9\sqrt{\ x}}$$

$$\frac{3\cos\left(x\right)\sqrt{\ x}}{9x}-\frac{3\sin\left(x\right)}{2\sqrt{\ x}}$$

$$3\cos\left(x\right)\sqrt{\ x}+\frac{3\sin\left(x\right)}{2\sqrt{\ x}}$$

$$f'(x)=-6x\sin(3x^2)$$

$$f'(x)=6x\cos(3x^2)$$

$$f'(x)=-3x\sin(3x^2)$$

$$f'(x)=3x\cos(3x^2)$$

$$f'(x)=2x\ln(x)+x$$

$$f'(x)=x^2\ln(x)+2x$$

$$f'(x)=2x\ln(x)-x$$

$$f'(x)=x^2\ln(x)-2x$$

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

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

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

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

$$f''(x)=2\sec^2(x)\tan(x)$$

$$f''(x)=\sec^2(x)$$

$$f''(x)=2\tan^2(x)$$

$$f''(x)=2\sec^4(x)$$

$$\frac{d^3y}{dx^3}=4xe^{\left(x^2+2\right)}\left(2x^2+3\right)$$

$$\frac{d^3y}{dx^3}=4x^3e^{\left(x^2+2\right)}$$

$$\frac{d^3y}{dx^3}=e^{\left(x^2+2\right)\left(2x+3\right)}$$

Not doing it this way, takes too long. Mr. Clarke will show me the shortcut shortly..