Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.
x = t − t\(^{−1}\), y = 6 + t\(^2\), t = 1
The correct answer needs to be in the form of y=something... I have this (incorrect):
\[y=(\large\frac{2t^3}{t^2+1})x+7\]
@ t=1, x=0 and y=7
\(\large\frac{dy}{dt}\)=2t
\(\large\frac{dx}{dt}\)= 1+\(\large\frac{1}{t^2}\) = \(\large\frac{t^2+1}{t^2}\)
Dividing by something, same as the product of its reciprocal, when doing \(\frac{dy}{dx}\)
Where's the error?

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It's looking correct to me...

except that you can plug in the value for t...
which just makes y=x+7

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