find an equation of the tangent line to the graph of y = (ln x)^2, at x = 3

Answer 1

the 1st derivative is the slope of the tangent line

Answer 2

y = ln² ( x )

Put x = 3 into the equation y = ln² ( x )

y ( 3 ) = ln² ( 3 )

y´( x ) = dy / dx = 2 ln ( x ) / x

Slope:

m ( x ) = y´( x ) = 2 ln ( x ) / x

Slope of y = ln² ( x ) at x = 3

m = y´( 3 ) = 2 ln ( 3 ) / 3

Equation of the line y = m x + b , for slope m = 2 ln ( 3 ) / 3, passing [ 3 , ln² ( 3 ) ]:

y = m x + b

ln² ( 3 ) = [ 2 ln ( 3 ) / 3 ] ∙ 3 + b

ln² ( 3 ) = 2 ln ( 3 ) + b

ln² ( 3 ) - 2 ln ( 3 ) = b

b = ln² ( 3 ) - 2 ln ( 3 )

y = m x + b

y = [ 2 ln ( 3 ) / 3 ] ∙ x + ln² ( 3 ) - 2 ln ( 3 )

Tangent line to y = ln² ( x ) at x = 3:

y = 2 x ∙ ln ( 3 ) / 3 + ln² ( 3 ) - 2 ln ( 3 )