Science:Math Exam Resources/Courses/MATH100/December 2016/Question 05 (b)

MATH100 December 2016

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Question 05 (b)
Let $x,y$ satisfy the equation $x^{y}=5y-x+1$. Find ${\frac {dy}{dx}}$ at the point $(x,y)=\left(1,{\frac {1}{5}}\right)$.

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
Use implicit differentiation.

Hint 2
To get the derivative of $x^{y}$, consider using logarithmic 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. Apply ${\frac {d}{dx}}$ to both sides of the equation $x^{y}=5y-x+1$. Differentiating the left-hand side requires work. We use logarithmic differentiation. Since $\log(x^{y})=y\log x$, we get the derivative of $x^{y}$ as ${\frac {d}{dx}}(x^{y})=x^{y}{\frac {d}{dx}}\log(x^{y})=x^{y}{\frac {d}{dx}}(y\log x)=x^{y}\left(y'\log x+{\frac {y}{x}}\right).$ This implies that $x^{y}\left(y'\log x+{\frac {y}{x}}\right)={\frac {d}{dx}}(x^{y})={\frac {d}{dx}}(5y-x+1)=5y'-1.$ Now plugging $x=1,y={\frac {1}{5}},$ we have $1\cdot \left(y'\cdot 0+{\frac {\frac {1}{5}}{1}}\right)=5y'-1$. So $\left.{\frac {dy}{dx}}\right|_{x=1,y={\frac {1}{5}}}=\left.y'\right|_{x=1,y={\frac {1}{5}}}=\color {blue}{\frac {6}{25}}$.