Answered

Suppose the curve defined by the equation 3x^2 = xy^2 + 1. Answer the following.

ACalculate dy/dx . Explain its meaning.
BFind the tangent line to the curve at the point (1, 2)

please help me!

Verified Answer :

LammettHash

\(\dfrac{\mathrm d}{\mathrm dx}\left[3x^2\right]=\dfrac{\mathrm d}{\mathrm dx}\left[xy^2+1\right]\)
\(\implies 6x=y^2+2xy\dfrac{\mathrm dy}{\mathrm dx}\)
\(\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{6x-y^2}{2xy}\)

The tangent line to the curve at \((1,2)\) has slope \(\dfrac{6\times1-2^2}{2\times1\times2}=\dfrac12\). In point-slope form, the line's equation is

\(y-2=\dfrac12(x-1)\implies y=\dfrac12x+\dfrac32\)

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