# Thread: Chain rule w/ trigonometric function help

1. ## Chain rule w/ trigonometric function help

find the derivative:
$\frac{d}{dx}(\cot^4 t -\csc^4 t)$

What did i do wrong?
$4(-\csc^2 t)^3 - 4(-\csc t\cot t)^3$
$-4\csc^6 t + 4\csc^3 t\cot^3 t$

Answer at back of book: $4\cot t \csc^2 t$

Thx & happy new year folks

Chain rule:

$\frac{d}{dt}[cot^{4}(t)]=-4cot^{3}(t)csc^{2}(t)$

$\frac{d}{dt}[csc^{4}(t)]=-4csc^{3}(t)(csc(t)cot(t))=-4csc^{4}(t)cot(t)$

${-}4cot^{3}(t)csc^{2}(t)+4csc^{4}(t)cot(t)$

Factor:

$4cot(t)csc^{2}(t)\underbrace{(csc^{2}(t)-cot^{2}(t))}_{\text{equals 1}}$

$\boxed{4cot(t)csc^{2}(t)}$

Your chain rule is a bit off . . .

Find the derivative: . $f(x) \:=\:\cot^4\!t -\csc^4\!t$

Answer: . $4\cot t \csc^2\!t$

We have: . $f(x)\:=\:(\cot t)^4 - (\csc t)^4$

Then: . $f'(t) \:=\:4\cot^3\!t(-\csc^2t) - 4\csc^3\!t(-\csc t\cot t) \:= \:-4\csc^2\!t\cot^3\!t + 4\csc^4\!t\cot t$

$\text{Factor: }\;f'(t)\:=\:4\cot t\csc^2\!t\underbrace{(\csc^2\!t - \cot^2\!t)}_{\text{This equals 1}}$

Therefore: . $\boxed{f'(x) \:=\:4\cot t\csc^2t}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We could have simplified at the very beginning . . .

$f(t)\:=\:\cot^4t - \csc^4t \:=\:\underbrace{(\cot^2x - \csc^2t)}_{\text{This equals -1}}(\cot^2t + \csc^2t)$

We have: . $f(t) \:=\:-\cot^2t - \csc^2t$

Then: . $f'(t)\:=\:-2\cot t(-\csc^2t) - 2\csc t(-\csc t\cot t)$

. . . . . $f'(t) \:=\:2\cot t\csc^2\!t +2\cot t\csc^2\!t \:=\:\boxed{4\cot t\csc^2\!t}$