Calculus Derivatives, Curve Sketching, & Integrals

$$f'\left(x\right)=\frac{\cos x\left(3x^2-1\right)-\sin x\left(6x\right)}{\left(3x^2-1\right)^2}$$

$$f'\left(x\right)=\frac{\sin x\left(6x\right)-\cos x\left(3x^2-1\right)}{\left(3x^2-1\right)^2}$$

$$f'\left(x\right)=\frac{\sin x\left(6x\right)+\cos x\left(3x^2-1\right)}{\left(3x^2-1\right)^2}$$

$$f'\left(x\right)=\frac{\cos x\left(3x^2-1\right)-\sin x\left(3x^2\right)}{\left(3x^2-1\right)^2}$$

$$g'\left(x\right)=\frac{12x^2-8}{\sqrt[]{2x^3-4x}}$$

$$g'\left(x\right)=\frac{3x^2-2}{\sqrt[]{2x^3-4x}}$$

$$g'\left(x\right)=\left(2x^3-4x\right)^{\frac{1}{2}}$$

$$g'\left(x\right)=\frac{1}{2}\left(2x^3-4x\right)^{\frac{-1}{2}}$$

$$y=9x+4$$

$$y=9x-4$$

$$y=\frac{11}{3}x-3$$

$$y=-\frac{3}{11}x-\frac{357}{11}$$

$$f'\left(x\right)=5\sec5x\csc5x$$

$$f'\left(x\right)=5\sec^2x$$

$$f'\left(x\right)=\frac{5}{\tan5x}$$

$$f'\left(x\right)=\frac{1}{5x}$$

 $f'\left(x\right)=e^{5x^2}$

 $f'\left(x\right)=5e^{5x^2}$

 $f'\left(x\right)=\frac{5}{2}xe^{x^2}$

 $f'\left(x\right)=10xe^{5x^2}$

$$\frac{13}{\left(5x+1\right)^2}$$

$$\frac{-7}{\left(5x+1\right)^2}$$

$$\frac{30x-13}{\left(5x+1\right)^2}$$

$$\frac{30x-7}{\left(5x+1\right)^2}$$

$$-\frac{6x}{\sqrt[]{1-3x^4}}$$

$$-\frac{6x}{\sqrt[]{1-9x^4}}$$

$$-\frac{6x}{\sqrt[]{1+3x^4}}$$

$$-\frac{6x}{\sqrt[]{1+9x^4}}$$