Differentiation Rules: math test solutions, College – download the PDF

1 Product and quotient rules / 4 pts

(a) \( f'(x)=3(x^2+4)+(3x-1)(2x)=3x^2+12+6x^2-2x=9x^2-2x+12 \). (2 pts)

(b) \( g'(x)=\dfrac{2(x-3)-(2x+1)(1)}{(x-3)^2}=\dfrac{-7}{(x-3)^2} \). (2 pts)

2 Chain rule / 3 pts

(a) \( y'=6(x^3+2)^5\cdot 3x^2=18x^2(x^3+2)^5 \). (1.5 pts)

(b) Product and chain rules: \( y'=-3e^{-3x}\cos(2x)+e^{-3x}(-2\sin(2x))=-e^{-3x}\big(3\cos(2x)+2\sin(2x)\big) \). (1.5 pts)

3 A logarithmic quotient / 3 pts

(a) \( y'=\dfrac{\frac1x\cdot x^2-\ln x\cdot 2x}{x^4}=\dfrac{x-2x\ln x}{x^4}=\dfrac{1-2\ln x}{x^3} \). (2 pts)

(b) \( y'=0 \) when \( 1-2\ln x=0 \), so \( \ln x=\tfrac12 \) and \( x=\sqrt e \). (1 pt)

4 Implicit differentiation / 3 pts

Check: \( 27+27=54=6\cdot 9 \). Differentiating: \( 3x^2+3y^2y'=6y+6x\,y' \). (1 pt)

So \( y'(3y^2-6x)=6y-3x^2 \), giving \( y'=\dfrac{2y-x^2}{y^2-2x} \). (1 pt)

At \( (3,3) \): \( y'=\dfrac{6-9}{9-6}=-1 \), so the tangent is \( y-3=-(x-3) \), i.e. \( y=-x+6 \). (1 pt)

5 Logarithmic differentiation / 4 pts

(a) \( \ln y=3x\ln x \), so \( \dfrac{y'}{y}=3\ln x+3 \) and \( y'=3x^{3x}(\ln x+1) \). (2 pts) At \( x=1 \): \( y'(1)=3\cdot 1\cdot(0+1)=3 \). (1 pt)

(b) \( \dfrac{y'}{y}=\dfrac{4}{x+1}+\dfrac{1}{2(x-2)}-\dfrac{6}{3x-1} \); at \( x=3 \): \( 1+\tfrac12-\tfrac34=\tfrac34 \). (1 pt)

6 Higher derivatives and concavity / 3 pts

(a) \( f'(x)=e^{-x}-x\,e^{-x}=(1-x)e^{-x} \). (1 pt)

(b) \( f''(x)=-e^{-x}-(1-x)e^{-x}=(x-2)e^{-x} \). (1 pt)

(c) Since \( e^{-x}\gt 0 \), \( f''=0 \) only at \( x=2 \); \( f''\gt 0 \) for \( x\gt 2 \) (bends upward) and \( f''\lt 0 \) for \( x\lt 2 \). (1 pt)