You should remember what is the equation of the tangent line to the graph of the function, at a point `x =x_0` such that:

`y - f(x_0) = f'(x_0)(x - x_0)`

You should find the derivative of the function using the quotient rule and the chain rule, such that:

`(dy)/(dx) = f'(x) = ((15e^(4x))'*x - 15e^(4x)*(x)')/x^2`

`f'(x) =(15*4x*e^(4x) - 15e^(4x))/x^2`

Factoring out `15e^(4x)` such that:

`f'(x) = (15e^(4x)(4x - 1))/x^2`

You need to find f'(1), hence, you should substitute 1 for x in equation `f'(x) = (15e^(4x)(4x - 1))/x^2` such that:

`f'(1) = (15e^(4)(4 - 1))/1^2 => f'(1) = 45e^4`

`f(1) = 15e^4`

Hence, you may write the equation of the tangent line such that:

`y - 15e^4 = 45e^4(x - 1)`

You need to put the equation `y - 15e^4 = 45e^4(x - 1)` in the slope intercept form such that:

`y = 15e^4 + 45e^4*x - 45e^4`

`y = 45e^4*x - 30e^4`

**Hence, evaluating the equation of the tangent line yields `y = 45e^4*x - 30e^4` , where `m = 45e^4` and `b = - 30e^4` . **