anonymous
  • anonymous
Derivative of tanx²
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
But i dont get how its chain rule
anonymous
  • anonymous
its tanx² not (tanx)² :S
Jack1
  • 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 \[u = g(x) = x ^{2}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Jack1
  • Jack1
ooo shiny, cheers
Jack1
  • Jack1
so first of all, how are your derivative skills? whats the derivative of g(x) above...?
Jack1
  • Jack1
@burhan101 ... whats the derivative of u... ? (thats the g(x) one)...
anonymous
  • anonymous
2u
Jack1
  • 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 ;)
Jack1
  • Jack1
cool, next we do the derivative of tan (u) (answwer in this case will contain a U)
Jack1
  • Jack1
so derivative of f(u) =...?
Jack1
  • 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 \[y' = 2x \times \sec ^{2}(x ^{2})\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.