quiz 4 - requires respondus lockdown browser + webcam\nstarted: sep 15 at 2:41pm\nquiz instructions\naccess…

quiz 4 - requires respondus lockdown browser + webcam\nstarted: sep 15 at 2:41pm\nquiz instructions\naccess code: start\ntimed: 30 minutes\nnumber of attempts: 1\nlockdown browser and respondus monitor required.\nthis quiz covers material up through week 4, where section 2.3 was covered. you may still need to know material weeks.\nquestion 4\nfind the derivative of (f(x)=\frac{qx + r}{sx + t}).\n(\frac{qt+rs}{(sx + t)^2})\n(\frac{qt - rs}{(sx - t)^2})\n(\frac{rs - qt}{(sx + t)^2})\nthe correct answer is not listed.\n(\frac{qt - rs}{(sx + t)^2})

quiz 4 - requires respondus lockdown browser + webcam\nstarted: sep 15 at 2:41pm\nquiz instructions\naccess code: start\ntimed: 30 minutes\nnumber of attempts: 1\nlockdown browser and respondus monitor required.\nthis quiz covers material up through week 4, where section 2.3 was covered. you may still need to know material weeks.\nquestion 4\nfind the derivative of (f(x)=\frac{qx + r}{sx + t}).\n(\frac{qt+rs}{(sx + t)^2})\n(\frac{qt - rs}{(sx - t)^2})\n(\frac{rs - qt}{(sx + t)^2})\nthe correct answer is not listed.\n(\frac{qt - rs}{(sx + t)^2})

Answer

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. For $f(x)=\frac{qx + r}{sx + t}$, let $u = qx + r$ and $v=sx + t$.

Step2: Find $u^\prime$ and $v^\prime$

Differentiate $u$ with respect to $x$: $u^\prime=\frac{d}{dx}(qx + r)=q$. Differentiate $v$ with respect to $x$: $v^\prime=\frac{d}{dx}(sx + t)=s$.

Step3: Apply quotient - rule

$f^\prime(x)=\frac{u^\prime v - uv^\prime}{v^{2}}=\frac{q(sx + t)-(qx + r)s}{(sx + t)^{2}}$.

Step4: Expand and simplify

Expand the numerator: $q(sx + t)-(qx + r)s=qsx+qt - qsx-rs=qt - rs$. So $f^\prime(x)=\frac{qt - rs}{(sx + t)^{2}}$.

Answer:

E. $\frac{qt - rs}{(sx + t)^{2}}$