How do you use implicit differentiation to find an equation of the tangent line to the curve #x^2 + 2xy − y^2 + x = 39# at the given point (5, 9)?

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Answer 1 · 2018-03-03T15:09:20

#29x-8y=73#

Differentiate the equation with respect to #x#:

#d/dx (x^2+2xy-y^2+x) = 0#

#2x +2y+2xdy/dx -2ydy/dx +1 = 0#

#(2x-2y) dy/dx = -1-2x-2y#

# dy/dx = (1+2x+2y)/(2y-2x)#

The equation of the tangent line is:

#y= y_0 + y'(x_0)(x-x_0)#

where #x_0 = 5#, #y_0 = 9# and:

#y'(x_0) = (1+10+18)/(18-10) = 29/8#

then:

#y = 9+29/8(x-5)#

#y=29/8x -73/8#

or:

#29x-8y=73#

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Answer 2 · 2018-03-03T15:17:09

#29x-8y-73=0#
is the equation of the tangent

Given:
#x^2+2xy-y^2+x=39#
when x=5, and y=9
#x^2+2xy-y^2+x=5^2+2xx5xx9-9^2+5#
#=25+90-81+5=(25+5)+(90-81)#
#=30+9=39#
#x^2+2xy-y^2+x=39#
verified

We have
#x^2+2xy-y^2+x=39#
Differentiating both sides wrt x
#2x+2(xdy/dx+y)-2ydy/dx+1=0#
Substituting x=5 and y=9
#2xx5+2(5xxdy/dx+9)-2xx9dy/dx+1=0#

#10+10dy/dx+18-18dy/dx+1=0#
#(10+18+1)+(10-18)dy/dx=0#
#(28+1)-8dy/dx=0#

#29=8dy/dx#

#dy/dx=29/8#
Slope of the tangent is
#m=29/8#
Passing through the point
#P-=(5,9)#
Equation of the tangent is
#(y-9)/(x-5)=29/8#

#8(y-9)=29(x-5)#
#8y-72=29x-145#
#27x-8y+72-145=0#

#29x-8y-73=0#