1. dy/dx of tan^2(x^3)

find $\frac{dy}{dx}$ of $\tan^2(x^3)$

I know that $\frac{dy}{dx}\tan{x} = \sec^2{x}$ and probably to use the chain rule
but didn't know how to deal with the powers?

the ans is: $6x^2\tan(x^3)\sec^2{(x^3)}$

2. Originally Posted by bigwave
find $\frac{dy}{dx}$ of $\tan^2(x^3)$

I know that $\frac{dy}{dx}\tan{x} = \sec^2{x}$ and probably to use the chain rule
but didn't know how to deal with the powers?

the ans is: $6x^2\tan(x^3)\sec^2{(x^3)}$
Note that $\tan^2(x^3)=(\tan(x^3))^2$.

Thus, when differentiate this, you will need to use chain rule twice.

Can you proceed?

3. next step

so if

then would my next step be:

$2(\tan{x^3})(\frac{-3x^2}{\cos^2{x^3}})$

not sure what you mean by doing chain rule twice

4. $y = (\tan x^3)^2$

Make $u = \tan x^3 \implies y = u^2 \implies \frac{dy}{du}=2u$

Applying the chain rule again separately on $u = \tan x^3$

Make $w = x^3 \implies u = \tan w \implies \frac{du}{dw}= \sec^2w \implies \frac{dw}{dx}= 3x^2$

$\frac{du}{dx}= \sec^2w \times 3x^2$

$\frac{dy}{dx} = 2u \times \sec^2w \times 3x^2 = 6x^2 \tan x^3 \sec^2 x^3$

5. Originally Posted by bigwave
so if

then would my next step be:

$2(\tan{x^3})(\frac{-3x^2}{\cos^2{x^3}})$ where did the minus come from, bigwave?

not sure what you mean by doing chain rule twice
$\frac{1}{cos^2\left(x^3\right)}=sec^2\left(x^3\rig ht)$

$x^3=u=f(x)$

$w=tanu=f(u)$

$y=w^2=f(w)$

$\frac{dy}{dx}=\frac{dy}{dw}\frac{dw}{du}\frac{du}{ dx}$

$=2wsec^2u3x^2=2[tanu]sec^2\left(x^3\right)3x^2=6x^2tan\left(x^3\right)s ec^2\left(x^3\right)$

Ah! pickslides! where'd you come from?

6. Originally Posted by Archie Meade
Ah! pickslides! where'd you come from?
Didn't mean to spoil the party...

7. someday I want be famous like you guys are. look a all those rep points..

a vain hope probably

the neg sign just slipped in by itself.... I certainly didn't put it there.

thnxs much... now I see the light... of u and w substitution