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Mathematics 25 Online
OpenStudy (anonymous):

Derivative of tanx²

OpenStudy (anonymous):

But i dont get how its chain rule

OpenStudy (anonymous):

its tanx² not (tanx)² :S

OpenStudy (jack1):

The chain rule can be applied to composites of more than two functions. consider : \[y = \tan (x ^{2})\] as: \[y = f(u) = \tan (u)\] and \[u = g(x) = x ^{2}\]

OpenStudy (jack1):

ooo shiny, cheers

OpenStudy (jack1):

so first of all, how are your derivative skills? whats the derivative of g(x) above...?

OpenStudy (jack1):

@burhan101 ... whats the derivative of u... ? (thats the g(x) one)...

OpenStudy (anonymous):

2u

OpenStudy (jack1):

yeah, pretty much... but in this case u refers to the function g(x) which is x^2 is derivative of u = 2x... not 2u, you got the right idea tho ;)

OpenStudy (jack1):

cool, next we do the derivative of tan (u) (answwer in this case will contain a U)

OpenStudy (jack1):

so derivative of f(u) =...?

OpenStudy (jack1):

k, if u come back, its a chain rule problem, so: differentiate x^2 (2x) then differentiate tan x^2 (derivative of tan = sec^2) and multiply your results \[y' = 2x \times \sec ^{2}(x ^{2})\]

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