Science:Math Exam Resources/Courses/MATH100/December 2010/Question 06

{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}

MATH100 December 2010

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q1 (l)  •  Q1 (m)  •  Q1 (n)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q4 (e)  •  Q5  •  Q6  •  Q7  •  Q8  •  

   Other MATH100 Exams

  •  December 2011  •  December 2010  •  December 2012  •  December 2013  •  December 2014  •  December 2015  •  December 2016  •  December 2018  •  

Question 06

Full-Solution Problems. In questions 2-8, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly requested. Find the derivative of the function ${\frac {t}{t+5}}$ using the definition of the derivative. NOTE: No credit will be given for using differentiation rules such as the power rule, product rule, quotient rule or chain rule, although you may use these rules to check your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The definition of a derivative is $\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$ Alternatively, one can use $\displaystyle f'(a)=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$

Hint 2
At some stage, you might want to find a common denominator for two fractions you are trying to subtract.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

By the definition of a derivative, we have that ${\begin{aligned}f'(t)&=\lim _{h\to 0}{\frac {f(t+h)-f(t)}{h}}\\&=\lim _{h\to 0}{\frac {{\frac {t+h}{t+h+5}}-{\frac {t}{t+5}}}{h}}\end{aligned}}$ The key step to simplifying this expression is here: we find a common denominator for the fractions on top and combine them. ${\begin{aligned}&=\lim _{h\to 0}{\frac {{\frac {(t+h)(t+5)}{(t+h+5)(t+5)}}-{\frac {t(t+h+5)}{(t+h+5)(t+5)}}}{h}}\\&=\lim _{h\to 0}{\frac {\frac {t^{2}+th+5t+5h-t^{2}-th-5t}{(t+h+5)(t+5)}}{h}}\end{aligned}}$ Now we simplify the expression into a single fraction and cancel the h on top and bottom. ${\begin{aligned}&=\lim _{h\to 0}{\frac {5h}{h(t+h+5)(t+5)}}\\&=\lim _{h\to 0}{\frac {5}{(t+h+5)(t+5)}}\end{aligned}}$ Finally taking the limit gives: ${\begin{aligned}&={\frac {5}{(t+5)(t+5)}}\\&={\frac {5}{(t+5)^{2}}}\end{aligned}}$

Solution 2

Using the alternative definition, ${\begin{aligned}f'(t)&=\lim _{x\to t}{\frac {f(x)-f(t)}{x-t}}\\f'(t)&=\lim _{x\to t}{\frac {{\frac {x}{x+5}}-{\frac {t}{t+5}}}{x-t}}\end{aligned}}$ Here we find a common denominator to simplify the fractions on top of the large fraction. ${\begin{aligned}f'(t)&=\lim _{x\to t}{\frac {{\frac {x(t+5)}{(x+5)(t+5)}}-{\frac {t(x+5)}{(t+5)(x+5)}}}{x-t}}\\f'(t)&=\lim _{x\to t}{\frac {\frac {xt+5x-tx-5t}{(x+5)(t+5)}}{x-t}}\end{aligned}}$ We then simplify the fraction, factoring out a 5 on top and then canceling the (x - t) term. ${\begin{aligned}f'(t)&=\lim _{x\to t}{\frac {5x-5t}{(x+5)(t+5)(x-t)}}\\f'(t)&=\lim _{x\to t}{\frac {5(x-t)}{(x+5)(t+5)(x-t)}}\\f'(t)&=\lim _{x\to t}{\frac {5}{(x+5)(t+5)}}\end{aligned}}$ Finally we take the limit. ${\begin{aligned}f'(t)&={\frac {5}{(t+5)(t+5)}}\\f'(t)&={\frac {5}{(t+5)^{2}}}\\\end{aligned}}$ Which completes the question.

{{#incat:MER CT flag||

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Limit definition of the derivative, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

}}