# Thread: Confunsing Chain Rules and more

1. Originally Posted by sk8erboyla2004
I really appreciate this bro, with each problem Im getting better with the last one you just posted i forgot to add the /g(x)^2 to the bottom

Thanks Ill have many more to come lol!

like this one

$r(p)=tan^3(4p)$

$r\prime(p)=\frac{4}{cos^3(4p)}$
You keep making mistakes because you're not doing all the steps.

Here $r = \tan^3{(4p)} = [\tan{(4p)}]^3$.

It's a function of a function, so use the chain rule.

Let $u = \tan{(4p)}$ so that $r = u^3$.

$\frac{du}{dp} = \frac{4}{\cos^2{(4p)}}$

$\frac{dr}{du} = 3u^2 = 3\tan^2{(4p)}$.

So $\frac{dr}{dp} = \frac{12\tan^2{(4p)}}{\cos^2{(4p)}}$.

2. Thanks

$h(t)=\arcsin(t^3)-ln(t^2+1)$

$h\prime(t)=\frac{3t^2}{\sqrt{1-t^6}}-\frac{2t}{t^2+1}$

?

$y=(csc(x))^20$

$y\prime=20(csc(x))^{19}*-csc(x)cot(x)$

3. Originally Posted by sk8erboyla2004
Thanks

$h(t)=\arcsin(t^3)-ln(t^2+1)$

$h\prime(t)=\frac{3t^2}{\sqrt{1-t^6}}-\frac{2t}{t^2+1}$

?

$y=(csc(x))^20$

$y\prime=20(csc(x))^{19}*-csc(x)cot(x)$
The first one looks ok...

For the second, remember $\csc{x} = \frac{1}{\sin{x}} = (\sin{x})^{-1}$.

So $y = (\csc{x})^{20} = (\sin{x})^{-20}$

Let $u = \sin{x}$ so that $y = u^{-20}$.

$\frac{du}{dx} = \cos{x}, \frac{dy}{du} = -20u^{-21} = -20(\sin{x})^{-21}$.

So $\frac{dy}{dx} = -20\cos{x}\sin^{-21}{x} = -20\cos{x}\csc^{21}{x}$.

4. Thanks mate! You really know your stuff the more you help the more I learn.

If you dont mind Im gonna keep this coming at for a good bit?

$y=-sin(9x^3+21)$

$y\prime=cos(9x^3+21)*(27x^2)$

? It seems im getting a tad better I think I have the chain rule down now I have to venture into derivatives of trig and inverse trig functions for this assignment

$y=sin(sin(x))$

[tex]y\prime=cos(sin(x))*cos(x)[/MATH

$y=cos(sec(x))$
$y\prime=-sin(sec(x))*sec(x)tan(x)$

Im sure you have any other what but is this way correct as well?

5. Originally Posted by sk8erboyla2004
Thanks mate! You really know your stuff the more you help the more I learn.

If you dont mind Im gonna keep this coming at for a good bit?

$y=-sin(9x^3+21)$

$y\prime=cos(9x^3+21)*(27x^2)$

? It seems im getting a tad better I think I have the chain rule down now I have to venture into derivatives of trig and inverse trig functions for this assignment

$y=sin(sin(x))$

[tex]y\prime=cos(sin(x))*cos(x)[/MATH

$y=cos(sec(x))$
$y\prime=-sin(sec(x))*sec(x)tan(x)$

Im sure you have any other what but is this way correct as well?
The first two are correct.

$y = \cos{\sec{x}} = \cos{\frac{1}{\cos{x}}} = \cos{[(\cos{x})^{-1}]}$.

Here you have a function of a function of a function, so you're going to have to use the chain rule twice.

Let $u = (\cos{x})^{-1}$ so that $y = \cos{u}$

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

$\frac{dy}{du} = -\sin{u} = -\sin{[(\cos{x})^{-1}]}$. That part was easy.

$\frac{du}{dx}$ is not quite so easy.

Let $v = \cos{x}$ so that $u = v^{-1}$.

$\frac{dv}{dx} = -\sin{x}$

$\frac{du}{dv} = -v^{-2} = -(\cos{x})^{-2}$.

So $\frac{du}{dx} = \sin{x}(\cos{x})^{-2}$.

Therefore

$\frac{dy}{dx} = -\sin{x}(\cos{x})^{-2}\sin{[(\cos{x})^{-1}]}=-\tan{x}\sec{x}\sin{(\sec{x})}$.

So that's right as well

6. Finally, Thanks man you were an amazing help!

Count this one solved for now. I just have good bit of problems left I wish I could get them checked but there are too many a las oh well, i feel confident now

Thanks Prove it reps and thanks coming your way!

Edit: Nvm I need a bit more checking lol

$y=e^{15csc(5x)}$
$y\prime=15csc(5)e^{15csc(5x)}*75csc(5x)cot(5x)$

7. Originally Posted by sk8erboyla2004
Finally, Thanks man you were an amazing help!

Count this one solved for now. I just have good bit of problems left I wish I could get them checked but there are too many a las oh well, i feel confident now

Thanks Prove it reps and thanks coming your way!

Edit: Nvm I need a bit more checking lol

$y=e^{15csc(5x)}$
$y\prime=15csc(5)e^{15csc(5x)}*75csc(5x)cot(5x)$
$y = e^{15\csc{(5x)}} = e^{15(\sin{(5x)})^{-1}}$.

Let $u = 15(\sin{(5x)})^{-1}$ so that $y = e^u$.

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

$\frac{dy}{du} = e^u = e^{15(\sin{(5x)})^{-1}}$

Let $v = \sin{(5x)}$ so that $u = 15v^{-1}$.

$\frac{dv}{dx} = 5\cos{(5x)}$

$\frac{du}{dv} = -15v^{-2} = -15(\sin{(5x)})^{-2}$

So $\frac{du}{dx} = -75\cos{(5x)}(\sin{(5x)})^{-2}$.

What does $\frac{dy}{dx}$ equal?

8. $e^{15csc(5x)}*-75csc(5x)cot(5x)$

?

9. Originally Posted by sk8erboyla2004
$e^{15csc(5x)}*-75csc(5x)cot(5x)$

?

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