Solution assignment 06 Implicit differentiation

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Assignment 6

Calculate the tangent line at the point  (1,3) of the graph of the following expression:

2x+3y^2=29

Solution

Implicit differentiation in combination with the chain rule yields:

2+6yy'=0, so:

y'=-\displaystyle\frac{1}{3y}

The derivative in (1,3) is:

y'=-\displaystyle\frac{1}{9}

The tangent line through the point (1,3) is:

y-3=-\displaystyle\frac{1}{9}(x-1)

or:

y=-\displaystyle\frac{1}{9}x+3\displaystyle\frac{1}{9}

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