Find the equation of the line tangent to y5y3xy4 at the poin

Find the equation of the line tangent to y^5+y^3=xy+4 at the point where y=1

the equation of the tangent line is y=

Solution

y^5+y^3=xy+4

at y =1

1^5+1^3=x*1 +4

1+1=x+4

x =-2

point on curve is (-2,1)

y^5+y^3=xy+4

differentiate with respect to x

5y⁴(dy/dx) +3y²(dy/dx)=y +x(dy/dx)+0

at point (-2,1)

5*1⁴(dy/dx) +3*1²(dy/dx)=1 -2(dy/dx)+0

8(dy/dx)=1-2(dy/dx)

10dy/dx =1

dy/dx =1/10

slope of tangent m = dy/dx =1/10

equation of tangent with slope m =1/10 at point (-2,1) is

y-1 =(1/10)(x-(-2))

y-1 =(1/10)(x+2)

y-1=(1/10)x +(2/10)

y=(1/10)x +(2/10)+1

y=(1/10)x +(12/10)

y=(1/10)x +(6/5)

equation of tangent is y=(1/10)x +(6/5)