Question
Use implicit differentiation to find the slope of the tangent line to the curve
y/(x–9y)=x^6–2
at the point (1,-1/-8).
Can someone please help me?
I don't understand
y/(x–9y)=x^6–2
at the point (1,-1/-8).
Can someone please help me?
I don't understand
Answers
Reiny
puzzled, why would your write the point as
(1 , -1/-8) and not as simply (1 , 1/8) ?
first expand it
y = x^7 - 2x - 9x^6 y + 18y
0 = x^7 - 2x - 9x^6 y + 17y
0 = 7x^6 - 2 - 9x^6 dy/dx - 54x^5y + 17dy/dx
dy/dx(17 - 9x^6) = 54x^5y - 7x^6
dy/dx = (54x^5y - 7x^6))/(17 - 9x^6)
at x = 1
dy/dx = (54(1/8) -7)/(17 - 9)
= (-1/4) / 8
= -1/32
equation of tangent:
y + 1/8 = (-1/32)(x-1)
y = (-1/32)x + 1/32 - 1/8
y = (-1/32)x -3/32
check my arithmetic
(1 , -1/-8) and not as simply (1 , 1/8) ?
first expand it
y = x^7 - 2x - 9x^6 y + 18y
0 = x^7 - 2x - 9x^6 y + 17y
0 = 7x^6 - 2 - 9x^6 dy/dx - 54x^5y + 17dy/dx
dy/dx(17 - 9x^6) = 54x^5y - 7x^6
dy/dx = (54x^5y - 7x^6))/(17 - 9x^6)
at x = 1
dy/dx = (54(1/8) -7)/(17 - 9)
= (-1/4) / 8
= -1/32
equation of tangent:
y + 1/8 = (-1/32)(x-1)
y = (-1/32)x + 1/32 - 1/8
y = (-1/32)x -3/32
check my arithmetic