The Chain Rule

  $$\frac{1}{3}$$

  $$\frac{2}{7}$$

$$\frac{1}{4}$$

$$\frac{3}{11}$$

$$\left(12x+4\right)\left(2x^2+4\right)^2$$  

$$3\left(2x^2+4\right)^2$$  

$$12x\left(2x^2+4\right)^2$$  

$$12\left(2x^2+4\right)^2$$  

$$\frac{3x^2+9}{2\ \sqrt[]{x^3+9x}}$$

$$\frac{x^2+9}{2\ \sqrt[]{x^3+9x}}$$

$$\frac{3x^2+9x}{2\ \sqrt[]{x^3+9x}}$$

$$\frac{3x^3+9x}{2\ \sqrt[]{x^3+9x}}$$

$$\frac{3x^2+9}{2\ \sqrt[]{2x^3+9}}$$

$$g'\left(x\right)=5\left(x^2+1\right)^4+2x$$  

$$g'\left(x\right)=5\left(x^2+1\right)^4$$  

$$g'\left(x\right)=5\left(2x\right)^4$$  

$$g'\left(x\right)=10x\left(x^2+1\right)^4$$  

$$5\left(x+3\right)^4$$  

$$15\left(x+3\right)^4$$  

$$5x\left(x+3\right)^4$$  

$$15x\left(x+3\right)^4$$  

$$3\left(2x-1\right)^2$$  

$$6\left(2x-1\right)^2$$  

$$6x\left(2x-1\right)^2$$  

$$2\left(2x-1\right)^3$$  

$$7\left(8-x\right)^6$$  

$$-7\left(8-x\right)^6$$  

$$-7x\left(8-x\right)^6$$  

$$7\left(8-x\right)^7$$  

$$36\left(3x+4\right)^5$$  

$$6\left(3x+4\right)^5$$  

$$12\left(3x+4\right)^5$$  

$$12x\left(3x+4\right)^5$$  

$$20\left(6-5x\right)^3$$  

$$4\left(6-5x\right)^3$$  

$$-20\left(6-5x\right)^3$$  

$$-5\left(6-5x\right)^3$$  

$$2(4x−1)^{-\frac{1}{2}}$$  

$$2(4x−1)^{\frac{1}{2}}$$  

$$4(4x−1)^{-\frac{1}{2}}$$  

$$4(4x−1)^{\frac{1}{2}}$$  

$$3\left(\frac{x}{2x+1}\right)^2$$  

$$3\left[\frac{1}{\left(2x+1\right)^2}\right]^2$$  

$$\frac{1}{2}$$  

$$\frac{3x^2}{\left(2x+1\right)^4}$$