# What is the equation of the line normal to # f(x)=-xln(2x^3-x)# at # x=1#?

##### 1 Answer

#### Explanation:

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point. The normal is perpendicular to the tangent and so the product of their gradients is

so If

# \ \ \ \ \ f'(x) = (-x)(1/(2x^3 - x)*(6x^2 - 1)) + (-1)(ln(2x^3 - x)) #

# :. f'(x) = -(6x^2-1)/(2x^2-1) -ln(2x^3-x) #

When

and

So the tangent passes through

# y-0 = 1/5(x-1) #

# \ \ :. y = 1/5x - 1/5 #

We can confirm this solution is correct graphically: