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:

$y-1=3(x-0)$

$y=3x+1$

Return to Assignments Implicit differentiation

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Solution assignment 10 Implicit differentiation

Assignment 10

Solution

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