What is the derivative of this

find derivative of [(sec x)^3 + (sec x)(tan x)^2 ]

By chain rule d/dx f(g(x)) = f'(g(x)) and product rule d/dx fg = f'g + fg', noting that d/dx sec(x) = sec(x)tan(x) and d/dx tan(x) = sec²(x), we obtain:

dy/dx = 3sec²(x)[sec(x)tan(x)] + sec(x)(2)(tan(x))(sec²(x)) + sec(x)tan(x)(tan²(x))
= 3sec³(x)tan(x) + 2sec³(x)tan(x) + sec(x)tan³(x)
= sec(x)tan(x)(3sec²(x) + 2sec²(x) + tan²(x))
= sec(x)tan(x)(5sec²(x) + tan²(x))

* Walkthrough *

For sec³(x), you apply chain rule, letting:

g(x) = sec(x)
g'(x) = sec(x)tan(x)
f(g(x)) = g(x)³
f'(g(x)) = 3g(x)²

For sec(x)tan²(x), apply both chain and product rules. Let:

f = sec(x)
f' = sec(x)tan(x)
g = tan²(x)
g' = 2tan(x)sec²(x)

There you go. That's how I differentiated the function!

I hope this helps!

Let y = (sec x)³ + (sec x)(tan x)²
Then let f(x) = (sec x)³ and g(x) = (sec x)(tan x)²

So we have y = f(x) + g(x)

We know y' = f'(x) + g'(x)
So we need to find f'(x) and g'(x) individually.

f'(x) can be found using chain rule:
Let u = sec x, u' = sec x tan x (can be found using quotient rule, by noticing sec x = 1/cos x)

f(u) = u³, f'(u) = 3u²

=> f'(x) = u' f'(u) = 3u² sec x tan x = 3 (sec³ x) (tan x)

Now we need to find g'(x). Use the product rule:
g'(x) = (sec x)' tan² x + (tan² x)' sec x

ASIDE: We can find (tan² x)' using chain rule: v = tan x, v' = sec² x
=> (tan² x)' = 2 tan x sec² x [END OF ASIDE]

Hence g'(x) = sec x tan³ x + 2 tan x sec³ x

Finally, y' = f'(x) + g'(x) = 3 (sec³ x) (tan x) + sec x tan³ x + 2 tan x sec³ x
= 5 (sec³ x) (tan x) + (sec x) (tan³ x)

sec(x)^3 + sec(x) * tan(x)^2 =>
sec(x)^3 + sec(x) * (sec(x)^2 - 1) =>
sec(x)^3 + sec(x)^3 - sec(x) =>
2 * sec(x)^3 - sec(x)
sec(x) * (2sec(x)^2 - 1)

Let sec(x) = u, then du = sec(x)tan(x)

u * (2u^2 - 1)

Take the derivative:

u * (4u * du) + (2u^2 - 1) * du
4u^2 * du + 2u^2 * du - du =>
du * (6u^2 - 1) =>
sec(x) * tan(x) * (6sec(x)^2 - 1)