Derivatives: Find these derivatives. Q18: $f(x) = 2x^3 - x^2 +3x$ Q36: $f(x)=3(5-x)^2$ Q38: $g(t) = 2 + 3 cos(t)$ Q42: $\displaystyle f(x) = x + {1\over x^2} = x + x^{-2}$ Q44: $\displaystyle h(x) = {2x^2-3x+1\over x}$ Q48: $f(x) = x^{1/3} + x ^{1/5}$ Q50: $f(t) = t^{2/3} - t^{1/3} +4$ Q56: $y = (x^2+2x)(x+1)$ Find the equation of the tangent at $(1,6)$.

QUESTION
Find the derivatives.

Q18: $f(x) = 2x^3 - x^2 +3x$ 
Q36: $f(x)=3(5-x)^2$ 
Q38: $g(t) = 2 + 3 cos(t)$ 
Q42: $\displaystyle f(x) = x + {1\over x^2} = x + x^{-2}$
Q44: $\displaystyle h(x) = {2x^2-3x+1\over x}$ 
Q48: $f(x) = x^{1/3} + x ^{1/5}$ 
Q50: $f(t) =  t^{2/3} - t^{1/3} +4$ 
Q56: $y = (x^2+2x)(x+1)$ 
         Find the equation of the tangent at $(1,6)$.  
 

SOLUTION

Basic derivative rule:    the derivative of $y=x^n$ is  $y'=nx^{n-1}$.

Q18: $f(x) = 2x^3 - x^2 +3x$ 
        The derivative is $f'(x)=6x^2-2x+3$ 

Q36: $f(x)=3(5-x)^2$ 
         $f'(x) = 3.2 (5-x)^1. (-1)$    the $-1$ is the derivative of $5-x$ (the inside function).                 
         $f'(x) = -6 (5-x)   or -30 +6x$ 

[ If you don't use this rule you can also just expand the brackets, simplify, then find derivative, as in $3(5-x)^2=3(25-10x+x^2)=75 - 30x + 3x^2$ and the derivative of this is $-30 +6x$ same as before] 
        
Q38: $g(t) = 2 + 3 \cos(t)$
        $g'(t) = 0 - 3 \sin(t) = -3 \sin(t)$
        . 
Q42: $f(x) =\displaystyle x + {1\over x^2} = x + x^{-2}$
         $f'(x) = 1 -2 x^{-3} =\displaystyle 1 - {2\over x^3}$
Q44: $h(x) = \displaystyle {2x^2-3x+1\over x}$ 
        $h(x) =\displaystyle {2x^2\over x} - {3x\over x}+ 1/x = 2x - 3 + x^{-1}$
        $\displaystyle h'(x) = 2 - 0 -1.x^{-2} =\displaystyle 2 - {1\over x^2}$

Q48: $f(x) = x^{1/3} + x ^{1/5}$
        $f'(x) = (1/3) x^{1/3 - 1} + (1/5) x ^{1/5-1 }$
        $f'(x) = (1/3) x^{-2/3} +(1/5)x^{-4/5}$
        $f'(x) = \displaystyle {1\over 3x^{2/3}} +{1\over 5x^{4/5}}$

Q50:  $f(t) =  t^{2/3} - t^{1/3} +4$
       $f'(t) = (2/3) t^{2/3-1} -(1/3) t^{1/3-1} + 0$
        $f'(t) = (2/3) t^{-1/3} - (1/3) t^{-2/3}$

Q56: $y = (x^2+2x)(x+1)$ 
        $y= x^3 + 2x^2 +x^2 + 2x$ (by expanding) 
        $y = x^3 + 3 x^2 +2x$ 
        $y' = 3x^2 + 6x + 2$ 
        At $(1,6)$, $y' = 3(1)^2 + 6(1) + 2 = 3 + 6 + 2 = 11$ 
       Tangent line is $y - 6 = 11 (x - 1 )$   (point gradient form of line) 
                            $y = 11x -11 + 6$ 
        Tangent Line:  $y = 11x - 5$