How do you determine dy/dx given x^(2/3)+y^(2/3)=a^(2/3)?
2 Answers
Explanation:
We will differentiate implicitly. This means, that whenever we differentiate a variable that's not
d/dxx^3=3x^2" " but" "d/dxy^3=3y^2dy/dx d/dxx=1" " but" "d/dxy=dy/dx d/dxxy^2=(d/dxx)y^2+x(d/dxy^2)=y^2+x(2y)dy/dx
So, for the given example, we see that we will use the power rule on the left-hand side. On the right, don't be misled by the two-thirds power:
d/dx(x^(2/3)+y^(2/3))=d/dxa^(2/3)
Power rule:
2/3x^(-1/3)+2/3y^(-1/3)dy/dx=0
Multiply both sides by
x^(-1/3)+y^(-1/3)dy/dx=0
1/x^(1/3)+1/y^(1/3)dy/dx=0
Solving for
1/y^(1/3)dy/dx=-1/x^(1/3)
dy/dx=-y^(1/3)/x^(1/3)=-(y/x)^(1/3)=-root3(y/x)
Explanation:
Rewrite it