eberechukwuk262 eberechukwuk262
  • 23-05-2024
  • Mathematics
contestada


Differentiation
Y = tan x³
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DavisLaurant
DavisLaurant DavisLaurant
  • 23-05-2024

The answer depends on if your problem is (tan x)^3 or tan(x^3)

You can u-sub for either:

(tanx)^3       u=tanx  du/dx=sec^2 x

u^3= y

dy/du=3u^2

dy/dx=3(tan x)^2 x sec^2 x

-------------------------------------------

y=tan (x^3)      u=x^3   du/dx= 3x^2

dy/du = sec^2 (u)

dy/dx = sec^2(x^3) x 3x^2

I hope this makes sense to you. Good luck!

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