Science:Math Exam Resources/Courses/MATH110/April 2018/Question 03 (a)

MATH110 April 2018

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Question 03 (a)
Find ${\textstyle y'}$ if ${\textstyle x^{3}-xy^{2}+y^{3}=1}$ .

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?

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Hint
Use implicit differentiation to differentiate each of the terms.

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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. We will use implicit differentiation to differentiate each of the terms and then we will put everything together. Let us start with the easy ones: we have ${\textstyle 1'=0}$ and ${\textstyle (x^{3})'=3x^{2}}$ . Using the product rule and then implicit differentiation, we have $(xy^{2})'=1\cdot y^{2}+x(y^{2})'=y^{2}+x\cdot 2y\cdot y'.$ For the last term we also use implicit differentiation to get ${\textstyle (y^{3})'=3y^{2}y'}$ . Putting everything together, we get $0=1'=(x^{3}-xy^{2}+y^{3})'=(x^{3})'-(xy^{2})'+(y^{3})'=3x^{2}-y^{2}-2xyy'+3y^{2}y'.$ Now we factor out the ${\textstyle y'}$ to get $3x^{2}-y^{2}+y'(3y^{2}-2xy)=0.$ Rearranging this equation gives $y'={\frac {y^{2}-3x^{2}}{3y^{2}-2xy}}.$ Answer: The answer is ${\color {blue}y'={\frac {y^{2}-3x^{2}}{3y^{2}-2xy}}}$ .