Solution assignment 10 Implicit differentiation

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Assignment 10

The equation:

x^2y^3+(y+1)e^{-x}=x+2

defines y as a differentiable function of x close to the point (x,y)=(0,1).

Calculate the tangent line of the graph at this point.

Solution

First we verify whether the point (0,1) lies on the graph of the expression, which appears to be the case.
Next we have to determine the derivative y' and do this by implicit differentiation in combination with the product and quotient rule:

2xy^3+x^2\cdot3y^2\cdot{y'}+y'\cdot{e^{-x}}+(y+1)\cdot-e^{-x}=1

For the point (0,1) we get y'=3

The equation of the tangent line through the point (0,1) of the graph is:

y-1=3(x-0)

y=3x+1

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