find the derivative of secx * ex
Answers
Answer:
[sec(x)]′
Lets use the definition of sec(x):
sec(x)=1cos(x)
[sec(x)]′=[1cos(x)]′
This is a rational function, so we have to use the quotient rule:
[1cos(x)]′=[1]′⋅cos(x)−1⋅[cos(x)]′(cos(x))2
=0⋅cos(x)−1⋅−sin(x)cos2(x)
=sin(x)cos2(x)
=sin(x)⋅1cos(x)⋅cos(x)
=sin(x)cos(x)⋅1cos(x)
Now use the identity:
sin(x)cos(x)=tan(x)
And:
1|cos(x)=sec(x)
So we will get:
=tan(x)⋅sec(x)=tan(x)sec(x)
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Kyle Sockalingum
, MMath Mathematics, University of St Andrews (2023)
Answered July 18, 2019
Once we recognise that secant is the reciprocal of cosine, we can find it’s derivative using the quotient rule:
y=secx=1cosx
dydx=0⋅cosx−(−sinx)⋅1cos2x
=sinxcos2x=1cosx⋅sinxcosx
I have written it as such because the first term becomes secant and the second tangent from elementary trigonometric identities. Therefore,
ddx[secx]=secxtanx
And we’re done.
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Manjunath Subramanya Iyer
, I am a retired bank officer teaching maths
Answered July 18, 2019
I am finding the derivative of sec x from first principles.
Let y = sec x …………….(1)
Let ∆x be a small change in x.
Let ∆ y be the corresponding change in y.
Then y + ∆y = sec ( x+ ∆ x) ……..(2)
(2) - (1) gives ∆ y = sec (x+ ∆x) - sec x
=> ∆ y = 1/{cos (x + ∆ x)} - 1/cos x
=> ∆y={ cos x - cos (x+∆ x)}/cos (x + ∆ x)}cos x.
∆y=[2 sin {x+ (∆x/2)}sin ∆x/2]/cos (x + ∆ x)}cos x
∆y/∆x = [{sin (x+(∆x/2)}{sin ∆x/2}/∆x/2]/cos (x + ∆ x)}cos x
As ∆x->0, sin ∆x/2}/∆x/2 = 1
So dy/dx = sin x / cos x cos x
=> dy/dx = (1/ cos x)( sin x /cos x)
=> dy/dx = sec x tan x
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How do I show that tan(X) / (sec(X) -1) = (sec(X) + 1)/tan(X)?
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What is the derivative of tan^-1 (sec(x) +tan(x))?
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Manju Jain
Answered July 18, 2019
Let y=sec(x)
y=1/cos(x). [ secx=1/ cosx]
Differentiating with respect to (x)
d(y)/dx=d(1/cosx)/dx
dy/dx=[cosx(0)-1(-sinx)]÷[cosx]^2
[ Using quotient rule if z=x/y
Then dz/dx= {dx/dx(y)-(x)dy/dx]÷(y^2)}]
Now, dy/dx=sin(x)/[cos(x)]^2
dy/dx=sinx / [cosx.cosx]
dy/dx=[sinx/cosx]×[1/cosx]
dy/dx=tanx.secx
[sinx / cosx = tanx and 1/cosx = secx]
dy/ dx = secx .tanx
Hence the derivative of sec(x) = (secx.tanx).
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Nicholas McConnell
, studied Mathematics at Rutgers University (2021)
Answered August 4, 2019
ddxsecx=ddx1cosx=−(cosx)′cos2x=sinxcos2x=1cosxsinxcosx=secxtanx
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Blake Winfield
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David Kaplan
, Ph.D. Mathematics, University of South Florida
Answered August 4, 2019
Rewrite as (cos(x))^(-1) and differentiate using the chain rule to get:
-1*(cos(x))^(-2)*(-sinn(xll
=
sin(xl / cos(x) * 1 / cos(xl
=
tan(x) * sec (x),
usually written as
sec*x) * tan(x)
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Blake Winfield
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Adel Alkhayat
, B.Sc. Mathematics, University of Basrah (1971)
Answered August 15, 2019
If you know the derivative of cos(x), then you can find the derivative of sec(x) = 1/cos(x).
Let f(x) = sec(x) = 1/cos(x) = [cos(x)]^-1, therefore:
f’(x) = - [cos(x)]^-2 . [- sin(x)] = [sin(x)/cos(x)][1/cos(x)] = tan(x) sec(x)
104 views · Answer requested by
Emily Cox
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