Question

What is the equation of the tangent line of `f(x) =x^2ln(1/x)-x` at `x = 4`?

Answer 1

`y =(-5-8ln4)x+16+16ln4`

Explanation:

We need to find the value of the gradient when `x=4` so we need to differentiate the expression (using the product rule):

So we need to know that `d/dx(uv) = (u)(d/dxv) + (d/dxu)(v)` and that `d/dx(lnx)=1/x`

So, `f(x)=x^2ln(1/x)-x`

`:. f(x)=x^2ln(x^-1)-x`
`:. f(x)=-x^2ln(x)-x`

Differentiating wrt `x` gives:
`f'(x)=(-x^2)(d/dxlnx)+d/dx(-x^2)(lnx)-1`

`=(-x^2)(1/x)+(-2x)(lnx)-1`
`=-x-2xlnx-1`

So, when `x=4 => f(4)=-16ln4-4`
And, `x=4 => f'(4) = -4-8ln4-1 = -5-8ln4`

Thus, the tangent line has slope `m=-5-8ln4` and passes through `(4,-16ln4-4)`, so using `y-y_1=m(x-x_1)` the equation of the tangent is given by:

`y - (-16ln4-4)=(-5-8ln4)(x-4)`
`:. y +16ln4+4=(-5-8ln4)x-4(-5-8ln4)`
`:. y +16ln4+4=(-5-8ln4)x+20+32ln4`
`:. y =(-5-8ln4)x+16+16ln4`