# Recitation 6

Exercise: Find the first and second derivatives.

1. $y=6x^2-10x-5x^{-2}$
2. $r=\frac{12}{\theta}-\frac{4}{\theta^3}+\frac{1}{\theta^4}$
1. $y'=12x-10+10x^{-3}$ and $y''=12-30x^{-4}$.
2. $r'=\frac{-12}{\theta^2}+\frac{12}{\theta^4}-\frac{4}{\theta^5}$ 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.

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

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.