Find The Equation of the Tangent to the Curve yln5x4 at the

Find The Equation of the Tangent to the Curve y=ln(5x+4) at the point where x=3.

Given that

y = ln(5x + 4) at the point x = 3

slope m = dy/dx

dy/dx = d/dx( ln(5x + 4) )

= 1/(5x+4) .d/dx(5x+4)

= 1/(5x+4).(5+0) [since,d/dx(ln(x)) = 1/x , d/dx(constant) = 0 ]

dy/dx = 5/(5x+4)

dy/dx at x = 3

dy/dx = 5/(5x+4)

dy/dx = 5/(5.3+4)

= 5/(15+4)

= 5/19

Hence,

slope m = 5/19

Substitute x = 3 in y

y = ln(5x+4)

= ln(5(3)+4)

= ln(19)

y = 2.94

Therefore,

( x , y ) = ( 3 , 2.94)

let ( x₁ , y₁ ) = ( 3 , 2.94 )

The equation of tangent line is ,

y - y₁ = m ( x - x₁ )

y - (2.94) = (5/19) ( x - 3 )

y - 2.94 = ( 5/19)(x-3)

y - 2.94 = (5x - 15)/19

y = (5x - 15)/19 + 2.94

y = ( 5x-15 + 55.86 )/19

y = ( 5x + 40.86) / 19

Therefore,

The equation of tangent is , y = ( 5x + 40.86) / 19