# Differentiation of sec^3( )

## Homework Statement

hello
i have to derive this equation
## sec^3(x^3+csc^-1(cot^3\frac{x+4}{3x})) ##

2. The attempt at a solution

## 3sectan^2(x^3+csc^-1(cot^3\frac{x+4}{3x}))*(3x^2-\frac{1}{|cot^3\frac{x+4}{3x}|\sqrt(cot^6\frac{x+4}{3x}-1)})*(-3csc^2\frac{x+4}{3x})*(\frac{3x*1-(x+4)(3)}{9x^2})##

is it so far right?

Mark44
Mentor

## Homework Statement

hello
i have to derive this equation
## sec^3(x^3+csc^-1(cot^3\frac{x+4}{3x})) ##
First off, it's not an equation. Second, you are differentiating the expression, not deriving it. Starting with the equation ax2 + bx + c = 0, I can derive the quadratic formula, and this has nothing to do with calculus differentiation.
Pual Black said:
2. The attempt at a solution

## 3sectan^2(x^3+csc^-1(cot^3\frac{x+4}{3x}))*(3x^2-\frac{1}{|cot^3\frac{x+4}{3x}|\sqrt(cot^6\frac{x+4}{3x}-1)})*(-3csc^2\frac{x+4}{3x})*(\frac{3x*1-(x+4)(3)}{9x^2})##

is it so far right?
Doesn't look right to me. It should start off with 3sec2(<bunch of other stuff>).

BTW, problems involving calculus should be posted in the Calculus section, not the Precalc section. I have moved this thread.

BiGyElLoWhAt
Gold Member
Looks like you have a lot of chain rule to apply. first off make the substitution u = sec(junk inside), then du =? plugging it in ##\frac{d}{dx}f(u) = \frac{df}{du}\frac{du}{dx}## keep making substitutions until you get rid of all the composites. for example, after you make the intial substitution, you need to make another one, composed within u, say v = junk inside, then dv = ? then within v, you have another etc...