Science:Math Exam Resources/Courses/MATH100/December 2015/Question 03 (ii)

MATH100 December 2015

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Question 03 (ii)
Let $xy^{2}+yx^{2}=2$. Find ${\frac {dy}{dx}}$ at the point (1,1). $-1$ $-1/3$ $0$ $1/3$ $1$ $2$

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
Try to use implicit differentiation.

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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. As we mention in the Hint, here we use the method of implicit differentiation. First differentiate both side of the given equation : ${\begin{aligned}{\frac {d}{dx}}&(xy^{2}+yx^{2})={\frac {d}{dx}}(2)\\&\implies {\color {OrangeRed}{\frac {d}{dx}}(xy^{2})}+{\color {OliveGreen}{\frac {d}{dx}}(yx^{2})}=0.\end{aligned}}$ Considering $y$ is a function of $x$, by the product rule and chain rule, we have ${\begin{aligned}{\color {OrangeRed}{\frac {d}{dx}}(xy^{2})}&=y^{2}{\frac {d}{dx}}(x)+x{\frac {d}{dx}}(y^{2})=y^{2}\cdot 1+x\cdot 2y{\frac {dy}{dx}}={\color {OrangeRed}y^{2}+2xy{\frac {dy}{dx}}}\\{\color {OliveGreen}{\frac {d}{dx}}(yx^{2})}&=x^{2}{\frac {dy}{dx}}+y{\frac {d}{dx}}(x^{2})=x^{2}{\frac {dy}{dx}}+y\cdot 2x={\color {OliveGreen}x^{2}{\frac {dy}{dx}}+2xy}.\end{aligned}}$ Plugging back these and solving the equation for ${\frac {dy}{dx}}$, ${\color {OrangeRed}y^{2}+2xy{\frac {dy}{dx}}}+{\color {OliveGreen}x^{2}{\frac {dy}{dx}}+2xy}=0$ and hence ${\frac {dy}{dx}}=-{\frac {y^{2}+2xy}{(2xy+x^{2})}}.$ Finally, by plugging $(x,y)=(1,1)$, we have ${\frac {dy}{dx}}|_{(x,y)=(1,1)}=-{\frac {3}{3}}={\color {blue}-1}.$ Therefore, the final answer is A.