Give the indicated derivative : `(d^3(tan ))/dx^3`  

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thilina-g's profile pic

thilina-g | College Teacher | (Level 1) Educator

Posted on

y = tanx

`(dy)/(dx) = sec^2(x)`

`(d^2y)/(dx^2) = 2*sec(x)*sec(x)*tan(x)`

`(d^2y)/(dx^2) = 2*sec^2(x)*tan(x)`

`(d^3y)/(dy^3) = 2(2*sec(x)*sec(x)*tan(x)*tan(x)+sec^2(x)*sec^2(x))`

` ` `(d^3y)/(dy^3) =2*sec^2(x)*(2*tan^2(x)+sec^2(x))`

but,

`1+tan^2(x) = sec^2(x)`

`(d^3y)/(dy^3) =2*sec^2(x)*(2*(sec^2(x)-1)+sec^2(x))`

`(d^3y)/(dy^3) =2*sec^2(x)*(3*sec^2(x) - 2)`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The third derivative of f(x) = tan x has to be determined.

f(x) = tan x

f'(x) = `sec^2x`

f''(x) = `2*sec x*sec x*tan x` = `2*sec^2x*tan x`

f'''(x) = `2*sec x*2*sec x*tan x*tan x + 2*sec^2x*sec^2x`

=> `4*sec^2 x*tan^2x + 2*sec^4 x`

The third derivative of tan x is `4*sec^2 x*tan^2x + 2*sec^4 x`

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