How do you derive the derivative of sec^3x?  

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pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

`sec^3(x)=1/(cos^3(x))`

`=(cos(x))^(-3)`          (i)

we know

`d /dx (cos(x))=-sin(x)`

`d/dx (x^n)=nx^(n-1)`

differentiate (i) with respect to x, pply chain rule ,we have

`d/dx(sec^3(x))=d/dx((cos(x))^(-3))`

`=-3(cos(x))^(-3-1)d/dx(cos(x))`

`=-3(cos(x))^(-4)(-sin(x))`

`=(3sin(x))/(cos(x))^4`

`=3tan(x)sec^3(x)`

 

Note:

`{sin(x)/(cos(x).cos^3(x))=(sin(x)/cos(x))(1/(cos^3(x)))`

`=tan(x)sec^3(x)``}`

 

oldnick's profile pic

oldnick | (Level 1) Valedictorian

Posted on

I'tafunction og function:

`y=f[g(x)]`   

Analisys teachs us :

`y'=f'[g(x)]g'(x)`

So in this case :  `f(x)= g^3(x)`    and `g(x)= sec x= 1/cosx`

`f'(x)=3sec^2x g'(x)`   (1)

Now `g(x)` too is a function of function

`g(x)=g[h(x)]`   where:   `g(x)=1/h(x)`   and `h(x)=cosx`

thus  `g'[h(x)]=g'(x)h'(x)= (sinx)/(cos^2x)`  (2)

So that substituing (2) in (1):

`f'(x)= (3sec^2x sinx)/(cos^2x)=3sec^4x sinx`

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