Finding equation of the tangent to the curve

Finding equation of the tangent to the curve

Postby Sylus » Thu Oct 15, 2020 4:02 pm

I need to find the equation of the tangent to the curve xy2 + 3x -3y=6. at the point (2,1)

I know i need to differentiate implicitly but its getting confusing differentiating the xy2?
After differentiation then id set the function dy/dx to equal 0???
Sylus
 

Re: Finding equation of the tangent to the curve

Postby Guest » Fri Oct 16, 2020 11:30 am

I assume that the "y2" is supposed to be [tex]y^2[/tex] (a better notation is "y^2").
The function is [tex]xy^2+ 3x- 3y= 6[/tex]. To find the derivative of y without having to solve for y, use the "chain rule". The derivative of [tex]xy^2[/tex] with respect to x is [tex]1(y^2)+ x\frac{d(y^2)}{dy}\frac{dy}{dx}= y^2+ 2xy\frac{dy}{dx}=0[/tex]. Differentiating [tex]xy^2+ 3x- 3y= 6[/tex] with respect to x gives [tex]y^2+ 2xy\frac{dy}{dx}+ 3- 3\frac{dy}{dx}= 0[/tex].

Now solve for dy/dx: [tex](2xy- 3)\frac{dy}{dx}= -3- y^2[/tex]. [tex]\frac{dy}{dx}= \frac{y^2+ 3}{3- 2xy}[/tex].

Setting x= 2, y= 1, [tex]\frac{dy}{dx}= \frac{1^2+ 3}{3+ 2(1)(2)}= \frac{4}{7}[/tex].
Guest
 


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