Mathematics, 28.01.2020 23:12 chivitogomez2400
Find an equation tangent to the curve at the given point. 1. x²y+xy²=3x @ (-1,1)
Answers: 1
via implicit differentiation:![x^2y+xy^2=3x\implies\dfrac{\mathrm d}{\mathrm dx}[x^2y+xy^2]=\dfrac{\mathrm d}{\mathrm dx}[3x]](/tpl/images/0479/0774/dad39.png)
![\dfrac{\mathrm d}{\mathrm dx}[x^2y]+\dfrac{\mathrm d}{\mathrm dx}[xy^2]=3](/tpl/images/0479/0774/bb9ea.png)
![\left(\dfrac{\mathrm d}{\mathrm dx}[x^2]y+x^2\dfrac{\mathrm d}{\mathrm dx}[y]\right)+\left(\dfrac{\mathrm d}{\mathrm dx}[x]y^2+x\dfrac{\mathrm d}{\mathrm dx}[y^2]\right)=3](/tpl/images/0479/0774/b9887.png)
on the given curve is the slope of the line tangent to it at this point. So at (-1, 1), we get
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Find an equation tangent to the curve at the given point. 1. x²y+xy²=3x @ (-1,1)...
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