Science:Math Exam Resources/Courses/MATH100/December 2019/Question 2 (b)

MATH100 December 2019

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Question 2 (b)
Find $\displaystyle {\frac {dy}{dx}}$ at the point (x,y)=(1,0) if $\arctan(xy)=2^{y}-x$.

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
Whenever there is a variable as an exponent, take the natural logarithm followed by the exponential. In the case at hand $2^{y}=e^{ln(2^{y})}=e^{yln(2)}$.

Hint 2
Treating y as a function of x, take the derivative w.r.t. to x of both sides of the equation. As a result of the chain rule there will be some y' terms on both sides. Solve for them after evaluating at (x,y)=(1,0).

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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Whenever there is a variable as an exponent, take the natural logarithm followed by the exponential. In the case at hand $2^{y}=e^{ln(2^{y})}=e^{yln(2)}$. Treating y as a function of x, the derivative w.r.t. to x of the LHS of the equation must be the same as that one of the RHS: ${\frac {d}{dx}}(\arctan(xy))={\frac {1}{1+(xy)^{2}}}\cdot {\frac {d(xy)}{dx}}={\frac {1}{1+(xy)^{2}}}\cdot (y+xy').$ While the right hand side is ${\frac {d}{dx}}(e^{yln(2)}-x)=(ln(2)y')e^{yln(2)}-1.$ Therefore, ${\frac {1}{1+(xy)^{2}}}\cdot (y+xy')=(ln(2)y')e^{yln(2)}-1.$ If the question asked for ${\frac {dy}{dx}}$ as a function of x we would just solve for y' in the previous equation. However, since they are asking for the value of this function at (x,y)=(1,0), it's faster to evaluate directly in this expression and then solve for y'(x=1). Indeed, substituting these values above results in: $y'=ln(2)y'-1.$ Which yields $y'(x=1)={\frac {1}{ln(2)-1}}.$