A mathematical problem (trigonometrical identities and differentiation)?

Question

How would you use the derivative of cosx to prove that

d/dx (secx) = secxtanx

I know the derivative of cosx = -sinx, but I keep getting stuck with all my attempts at answering

Any help would be much appreciated!

Answer 1

A'ighty, first write sec(x) = 1/cos(x). Now take the derivative using the Quotient Rule:

d/dx (1/cos(x)) = (cos(x)*d/dx(1) - 1*d/dx(cos(x))) / cos^2(x)

= (0 - (-sin(x))) / cos^2(x)

= sin(x) / cos^2(x)

= (1/cos(x)) * (sin(x) / cos(x))

= sec(x)*tan(x).

And there you go. ^_^

Answer 2

Just observe that sec(x) = 1/cos(x) = cos(x)^(-1), whenever cos (x) <> 0. Now, by the chain rule, it follows that d/dx(secx(x) = (-1) * cos(x)^(-2) * d/dx(cos(x) = (-1) * (1/cos (x)) * (1/cos(x)) * (-sin(x) = sec (x) * sin(x)/cos(x). Since sin(x)/cos(x) = tan(x), we gfinally have that d/dx(sec(x)) = sec(x) * tan(x)

Answer 3

d(u/v)=(vdu-udv)/v^2...(1)
(quotient rule)

secx=1/cosx

let u=1,v=cosx

substitute into (1)

d(1/cosx)=(cosx*0-1*(-sinx))
/cosx*cosx
=(0 +sinx)/
cosx*cosx
=(sinx/cosx)*1/cosx
=tanx*secx
as required

i hope that this helps

Answer 4

(a million - cosA)/sinA + sinA/(a million - cosA) = 2cscA hint: Please determine you employ areas and parentheses to make your question sparkling. :) Yahoo!solutions truncates the question regardless of if this is basically too long and would not contain areas. Take the LHS. (a million - cosA)/sinA + sinA/(a million - cosA) = the effortless denominator is sinA(a million - cosA), so multiply as mandatory. [(a million - cosA)/sinA * (a million - cosA)/(a million - cosA)] + [sinA/(a million - cosA) * (sinA/sinA)] = [(a million - cosA)(a million - cosA)/sinA(a million - cosA)] + [sin²A/(a million - cosA)sinA] = Simplify. [(a million - 2cosA + cos²A)/sinA(a million - cosA)] + [sin²A/(a million - cosA)sinA] = (a million - 2cosA + cos²A+ sin²A) / (a million - cosA)sinA = remember that cos²A+ sin²A = a million (a million - 2cosA + a million) / (a million - cosA)sinA = (2 - 2cosA) / (a million - cosA)sinA = element. 2(a million - cosA) / (a million - cosA)sinA = The (a million - cosA) cancel. 2 / sinA = 2 * a million/sinA = remember that cscA = a million/sinA 2 * cscA = 2cscA = RHS desire this helped!

Answer 5

d/dx (sec x) =
d/dx (1 / cos x)

Quotient rule
(f/g)' = (f'g - fg') / (g²)

f(x) = 1; g(x) = cos x
f'(x) = 0; g'(x) = -sin x

d/dx (1 / cos x) = ( 0 - (-sin x) ) / cos² x
= sin x / cos² x
= (1 / cos x) (sin x / cos x)
= sec x tan x