Quotient rule and tangent

Practise quotient rule and tangent with an original VectorPaths question, worked explanation and related VCE Methods practice.

4 practice marks

Original practice question

Let $h(x)=(2x+1)/(x-1)$, $x\ne1$. (a) Find and simplify $h\prime(x)$. [2 marks] (b) Find the tangent at $x=2$. [2 marks]

Worked explanation

• Part (a)

  The quotient rule gives $[2(x-1)-(2x+1)]/(x-1)^2=-3/(x-1)^2$.

• Part (b)

  $h(2)=5$ and $h\prime(2)=-3$. Thus $y-5=-3(x-2)$, giving $y=-3x+11$.

(a) $-3/(x-1)^2$.; (b) $y=-3x+11$.

Original VectorPaths practice; not an official VCAA question or marking rubric.