Question

Using quotient rule find dy/dx of:

y=(4+x)/(cos(x))

and find the equation of the normal to the curve y=(4+x)/(cos(x))

at the pint where x=0

Use y=mx+c form

Answer 1

dy/dx = (cosx(1) - (4+x)(-sinx))/cos^2 x
when x = 0
dy/dx = (1(1) - 4(0))/1
= 1

so the slope of the tangent is 1
then the slope of the normal is -1

when x = 0, y = 4/1 = 4

so (y-4) = (-1)(x-0)
y = -x + 4

Answer 2

so, did you try the rule? why not show what you got?

If y = u/v, where u and v are functions of f, then

y' = (u'v - uv')/v^2
In this case, that means

y' = {(1)(cosx) - (4+x)(-sinx)]/(cosx)^2

At x=0,
y = (4+0)/(1) = 4
y' = [(1)(1)-(4)(0)]/1^2 = 1
So, the normal has slope -1

Now you have a point (0,4) and a slope (-1). Use the point-slope form for the line, then rearrange things to slope-intercept form.