Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 04 (a)

MATH 180 December 2017

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[hide] Question 04 (a)
(a) Let $\tan(x-y)={\frac {y}{1+x^{2}}}$. Calculate ${\frac {dy}{dx}}$ at ${\textstyle (0,0)}$.

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 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.

[show] Hint 1
Use implicit differentiation.

[show] Hint 2
Differentiate both sides with respect to ${\textstyle x}$, and treat $y$ as a function of $x$; ${\textstyle y=y(x)}$.

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[show] Solution

As in Hint 2, we differentiate both sides with respect to ${\textstyle x}$. That means we treat ${\textstyle y}$ as a function of ${\textstyle x}$, namely ${\textstyle y=y(x)}$. For the left hand side, we use the chain rule with ${\textstyle f(x)=\tan(x)}$, ${\textstyle g(x)=x-y(x)}$, ${\textstyle f'(x)=\sec ^{2}(x)}$, and ${\textstyle g'(x)=1-{\frac {dy}{dx}}}$ to yield $$(f(g(x)))'=f'(g(x))g'(x)=\sec ^{2}(x-y(x))\cdot \left(1-{\dfrac {dy}{dx}}\right).$$ For the right hand side we use the quotient rule with ${\textstyle F(x)=y(x)}$, ${\textstyle G(x)=1+x^{2}}$, ${\textstyle F'(x)={\dfrac {dy}{dx}}}$ and ${\textstyle G'(x)=2x}$ to yield $$\left({\dfrac {F(x)}{G(x)}}\right)'={\dfrac {F'(x)G(x)-F(x)G'(x)}{G(x)^{2}}}={\dfrac {{\frac {dy}{dx}}\cdot (1+x^{2})-y(x)\cdot (2x)}{(1+x^{2})^{2}}}.$$ Therefore, we have $$\sec ^{2}(x-y(x))\cdot \left(1-{\dfrac {dy}{dx}}\right)={\dfrac {(1+x^{2}){\frac {dy}{dx}}-2xy(x)}{(1+x^{2})^{2}}}.$$ To find ${\textstyle {\frac {dy}{dx}}}$ at ${\textstyle (0,0)}$, we put ${\textstyle x=0}$ and ${\textstyle y=0}$ in the above expression: $$\sec ^{2}(0-0)\cdot \left(1-{\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,0)}\right)={\dfrac {(1+0^{2}){\frac {dy}{dx}}{\bigg |}_{(x,y)=(0,0)}-2\cdot 0\cdot 0}{(1+0^{2})^{2}}}.$$ Since ${\textstyle \sec(0)=1}$, the above equation simplifies to $$1-{\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,0)}={\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,0)}$$ We solve this equation in the unknown ${\textstyle {\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,0)}}$: $$1=2{\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,0)}$$ $${\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,0)}={\dfrac {1}{2}}.$$ Answer: ${\textstyle {\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,0)}={\color {blue}{\dfrac {1}{2}}}}$.