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
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
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.
To find the derivative of the function y = (4 + x) / cos(x), we can use the quotient rule.
The quotient rule states that for two functions u and v, the derivative of their quotient is given by:
(dy/dx) = (v * du/dx - u * dv/dx) / (v^2)
Let's differentiate the function step-by-step:
1. Let u = (4 + x) and v = cos(x).
2. Find du/dx (the derivative of u) and dv/dx (the derivative of v):
- du/dx = d/dx (4 + x) = 1 (since the derivative of a constant is zero and the derivative of x is 1)
- dv/dx = d/dx(cos(x)) = -sin(x) (using the chain rule, where the derivative of cos(x) is -sin(x))
3. Substitute the values into the quotient rule formula:
(dy/dx) = (v * du/dx - u * dv/dx) / (v^2)
= (cos(x) * 1 - (4 + x) * (-sin(x))) / (cos(x))^2
= (cos(x) + (4 + x) * sin(x)) / cos^2(x)
So, the derivative of y = (4 + x) / cos(x) is:
(dy/dx) = (cos(x) + (4 + x) * sin(x)) / cos^2(x)
Now, let's find the equation of the normal to the curve at the point where x = 0. The normal to a curve is a line that is perpendicular to the tangent of the curve at the given point.
At x = 0, the value of y is:
y = (4 + 0) / cos(0) = 4 / 1 = 4
To find the slope of the tangent at x = 0, we can substitute x = 0 into the derivative we obtained earlier:
(dy/dx) = (cos(0) + (4 + 0) * sin(0)) / cos^2(0)
= (1 + 4 * 0) / 1^2 = 1
Since the slope of the tangent is 1, the slope of the normal will be the negative reciprocal of the tangent's slope. Therefore, the slope of the normal is -1.
Using the point-slope form of a line (y - y1 = m(x - x1)), we can plug in the values (x1, y1) = (0, 4) and m = -1:
y - 4 = -1(x - 0)
y - 4 = -x
y = -x + 4
Hence, the equation of the normal to the curve y = (4 + x) / cos(x) at the point where x = 0 is y = -x + 4.