**Exercise:** Find the first and second derivatives.

- y=6x^2-10x-5x^{-2}
- r=\frac{12}{\theta}-\frac{4}{\theta^3}+\frac{1}{\theta^4}

- y'=12x-10+10x^{-3} and y''=12-30x^{-4}.
- r'=\frac{-12}{\theta^2}+\frac{12}{\theta^4}-\frac{4}{\theta^5} and r''=\frac{24}{\theta^3}-\frac{48}{\theta^5}+\frac{20}{\theta^6}.

**Exercise:** Find y' (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate, where y=(2x+3)(5x^2-4x).

By Product Rule, y'=2(5x^2-4x)+(2x+3)(10x-4)=30x^2+14x-12. On the other hand, y=-12x+7x^2+10x^3. Hence y'=-12+14x+30x^2.

**Exercise:** Find the derivatives of the functions.

- y=\frac{2x+5}{3x-2}
- z=\frac{4-3x}{3x^2+x}
- v=(1-t)(1+t^2)^{-1}

- y'=\frac{2(3x-2)-3(2x+5)}{(3x-2)^2}=\frac{-19}{(3x-2)^2}.
- z'=\frac{-3(3x^2+x)-(4-3x)(6x+1)}{(3x^2+x)^2}=\frac{9x^2-24x-4}{x^2(3x+1)^2}.
- v'=(-1)(1+t^2)^{-1}+(1-t)(-1)(1+t^2)^{-2}=\frac{-t^2+t-2}{(1+t^2)^2}

**Exercise:** Find the equations of the tangent line and normal lien to the curve at the given point. y=x^2-x^4, (1, 0).

Calculate the derivative, y'=2x-4x^3. Evaluate at x=1, the slope of the tangent line is 2-4=-2. So the equation of the tangent line is y=-2x+1. On the other hand, the slope of the normal line has the slope \frac{1}{2}. So the equation of the normal line is y=\frac{1}{2}x+1.

**Exercise:** Show that the curve y=6x^3+5x-3 has no tangent line with slope 4.

Calculate y'=18x^2+5\geq 5. So the slope of any tangent line is always greater than 4.