Science:Math Exam Resources/Courses/MATH102/December 2011/Question 01 (b) i

MATH102 December 2011

  •  Q1 (a) i  •  Q1 (a) ii  •  Q1 (b) i  •  Q1 (b) ii  •  Q1 (b) iii  •  Q1 (c)  •  Q2 (a) i  •  Q2 (a) ii  •  Q2 (a) iii  •  Q2 (b) i  •  Q2 (b) ii  •  Q2 (b) iii  •  Q2 (c)  •  Q3  •  Q4 (i)  •  Q4 (ii)  •  Q4 (iii)  •  Q4 (iv)  •  Q4 (v)  •  Q4 (vi)  •  Q5  •  Q6  •  Q7 (i)  •  Q7 (ii)  •  Q7 (iii)  •  Q8 (i)  •  Q8 (ii)  •  Q8 (iii)  •  Q9  •  

   Other MATH102 Exams

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

Question 01 (b) i
Short-Answer Questions. A correct answer in the box gives full marks. For partial marks work needs to be shown. Consider the curve $\displaystyle e^{2x}+y(1-x)=1$ Find ${\frac {dy}{dx}}$ at the point x = 0.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
You can either use implicit differentiation or solve for y. Give it a try, we provide a solution for each method.

Checking a solution serves two purposes: helping you if, after having used the hint, 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To do this problem, we rearrange the equation to isolate $y$ and then take the derivative of $y$ with respect to $x$: ${\begin{aligned}e^{2x}+y(1-x)=1\\\rightarrow \quad y={\frac {1-e^{2x}}{1-x}}\end{aligned}}$ Using the quotient rule, we evaluate $dy/dx$ and we are done: ${\begin{aligned}{\frac {dy}{dx}}={\frac {(-2e^{2x})(1-x)-(-1)(1-e^{2x})}{(1-x)^{2}}}={\frac {-3e^{2x}+2xe^{2x}+1}{(x-1)^{2}}}.\end{aligned}}$ Hence, at x = 0 we obtain ${\frac {dy}{dx}}(0)={\frac {-3e^{0}+0+1}{(-1)^{2}}}=-2$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can also use implicit differentiation. By differentiating with respect to x we obtain: $2e^{2x}+{\frac {dy}{dx}}(1-x)+y(-1)=0$ (Watch out for the product rule on the term y(1-x)) The difference is that we need to know the value of y when x = 0, but if we plug that in the original equation we obtain $\displaystyle e^{0}+y(1-0)=1$ and so we have that y = 0. We use this in what we obtained by differentiating implicitly: $2e^{0}+{\frac {dy}{dx}}+0=0$ and so we have that ${\frac {dy}{dx}}(0)=-2$

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Implicit differentiation, MER Tag Quotient rule, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag