In the attached problem there are two parts I had to figure out. For part a) I had to find dy/dx in terms of the variable t using the information stated in the top. However, I'm not confident about my answer for part b). Can anyone check to see that I have answered part b) correctly? My answer for part b) is at the bottom right of the image. Thank you!
2 Answers
$\begingroup$Your work is correct except for a minus sign. In your final answer, the denominator is $x^3-3x^2+3x\mathbf{-}1$, not $x^3-3x^2+3x\mathbf{+}1$? The correct answer has a minus sign. I'll just note there's a slightly easier way.$$y=\frac{x^2}{x^2-2x+1}=\left(\frac{x}{x-1}\right)^2$$so$$y'= 2\frac{x}{x-1}\left(\frac{x}{x-1}\right)'$$
$\endgroup$ 7 $\begingroup$The answer for a) if you need it is simple. $\dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}.$ Substitute $\dfrac{dy}{dt} = 2t$ and $\dfrac{dx}{dt}= \dfrac{1}{(1+t)^2}.$
So we have $\dfrac{dy}{dx} = 2t(1+t)^2$ and $x=1-\dfrac{1}{1+t}.$ Then $1+t = \dfrac{1}{1-x}$ and $t=\dfrac{1}{1-x}-1.$ Just substitute the values into the expression for $\dfrac{dy}{dx}$ to get your answer.
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