Question

Find the slope of the line normal to the graph of f ( the square root of x^4-16x^2) at x=5

Answer 1

differentiate the function

\[dy/dx = 4x^3 - 32x\] substitute the value x = 5

this will give the gradient of the tangent (m). The gradient of the normal is -1/m

Answer 2

the equation was actually \[\sqrt{x^4-16x^2}\] which after i differentiated it i got 1/2 (x^4-16x^2) (4x^3-32x) so then i would just set find fprime of 5 using what i differentiated?

Answer 3

thats it... that will give the gradient of the tangent...... then -1/m for the normal