# Thread: Derivative of a Squared Trignometric Function

1. ## Derivative of a Squared Trignometric Function

I'm a little confused with this question:

If $y = sec^s (x^2)$, then $\frac {dy}{dx} =$

1. (A) $2xsec(x^2)tan(x^2)$
2. (B) $4xsec(x^2)tan(x^2)$
3. (C) $sec^2(x^2)tan(x^2)$
4. (D) $4xsec^2(x^2)$
5. (E) $4xsec^2(x^2)tan(x^2)$

See, what I'm not sure about is whether $sec^2(x^2)$ can be written as $(sec x)^2$ or $(sec x^2)^2$

In the first way, I'd get:

$
\frac {d}{dx}[(sec x)^2] = 2secxtanx$

which isn't a choice, but choice (A) is similar...

And if I use the second way:

$
\frac {d}{dx}[(sec x^2)^2] = 2(sec x^2)(sec x^2tan^2)
$

Which isn't in the choices, either. >_<

Help ? Please & Thank you =)

2. Originally Posted by StarlitxSunshine
And if I use the second way:

$
\frac {d}{dx}[(sec x^2)^2] = 2(sec x^2)(sec x^2tan^2)
$

Which isn't in the choices, either. >_<
It is if you allow the chain rule once more and multiply it all by 2x...

(because when you differentiated the inner function sec(x^2) to get sec(x^2)tan(x^2) you needed the chain rule for that too, because of the 'inner-inner' function x^2.)

3. Originally Posted by StarlitxSunshine
I'm a little confused with this question:

If $y = sec^s (x^2)$, then $\frac {dy}{dx} =$

1. (A) $2xsec(x^2)tan(x^2)$
2. (B) $4xsec(x^2)tan(x^2)$
3. (C) $sec^2(x^2)tan(x^2)$
4. (D) $4xsec^2(x^2)$
5. (E) $4xsec^2(x^2)tan(x^2)$

See, what I'm not sure about is whether $sec^2(x^2)$ can be written as $(sec x)^2$ or $(sec x^2)^2$

In the first way, I'd get:

$
\frac {d}{dx}[(sec x)^2] = 2secxtanx$

which isn't a choice, but choice (A) is similar...

And if I use the second way:

$
\frac {d}{dx}[(sec x^2)^2] = 2(sec x^2)(sec x^2tan^2)
$

Which isn't in the choices, either. >_<

Help ? Please & Thank you =)
$
sec^2(x^2) = [sec(x^2)][sec(x^2)] = [sec(x^2)]^2$

You'll need to use the chain rule to differentiate this

$2sec(x^2) \cdot 2x \cdot sec(x^2)tan(x^2) = 4xsec^2(x^2)tan(x^2)$ which corresponds to option E

4. Originally Posted by StarlitxSunshine
I'm a little confused with this question:

If $y = sec^s (x^2)$, then $\frac {dy}{dx} =$

1. (A) $2xsec(x^2)tan(x^2)$
2. (B) $4xsec(x^2)tan(x^2)$
3. (C) $sec^2(x^2)tan(x^2)$
4. (D) $4xsec^2(x^2)$
5. (E) $4xsec^2(x^2)tan(x^2)$

See, what I'm not sure about is whether $sec^2(x^2)$ can be written as $(sec x)^2$ or $(sec x^2)^2$

In the first way, I'd get:

$
\frac {d}{dx}[(sec x)^2] = 2secxtanx$

which isn't a choice, but choice (A) is similar...

And if I use the second way:

$
\frac {d}{dx}[(sec x^2)^2] = 2(sec x^2)(sec x^2tan^2)
$

Which isn't in the choices, either. >_<

Help ? Please & Thank you =)
$y = \sec ^2 (x^2)$

$y= (\sec (x^2))^2$

$\frac{dy}{dx} = 2 \left(\frac{d(\sec (x^2))}{dx} \right)(\sec (x^2) )^{2-1}$

$\frac{dy}{dx} = 2\left(\frac{d(x^2)}{dx}\right)(\sec (x^2)\tan (x^2))(sec(x^2))$

$\frac{dy}{dx} = 4x\sec ^2(x^2) \tan x^2$

5. The function $y=\sec^2(x^2)$ if the composition of 3 functions. This means you need to use the chain rule.

This can be seen as

$y=s^2 \implies \frac{dy}{ds}=2s$
$s=sec(t) \implies \frac{ds}{dt}=\sec(t)\tan(t)$
$t=x^2 \implies \frac{dt}{dx}=2x$

So the chain rule tells us the derivative is

$\frac{dy}{dx}=\frac{dy}{ds}\frac{ds}{dt}\frac{dt}{ dx}=2s(\sec(t)\tan(t))(2x)$

Now just sub in from above to get

$\frac{dy}{dx}=2\sec(t)\tan(t)(2x)=4x\sec(x^2)\tan( x^2)$

6. Ohh... I understand it now! It's like double chain rule :33

Thank you very much !

=)

7. Originally Posted by TheEmptySet
The function $y=\sec^2(x^2)$ if the composition of 3 functions. This means you need to use the chain rule.

This can be seen as

$y=s^2 \implies \frac{dy}{ds}=2s$
$s=sec(t) \implies \frac{ds}{dt}=\sec(t)\tan(t)$
$t=x^2 \implies \frac{dt}{dx}=2x$

So the chain rule tells us the derivative is

$\frac{dy}{dx}=\frac{dy}{ds}\frac{ds}{dt}\frac{dt}{ dx}=2s(\sec(t)\tan(t))(2x)$

Now just sub in from above to get

$\frac{dy}{dx}=2\sec(t)\tan(t)(2x)=4x^2\sec(x^2)\ta n(x^2)$
Wait... I didn't see your post when I posted:

I don't understand the very last step.

How did you go from:

$\frac{dy}{dx}=2\sec(t)\tan(t)(2x)$

8. Originally Posted by StarlitxSunshine
Wait... I didn't see your post when I posted:

I don't understand the very last step.

How did you go from:

$\frac{dy}{dx}=2\sec(t)\tan(t)(2x)$
He/she defined $t= x^2$ earlier in their post so it was subbing back x^2 for t

9. Oh..

Umm...But that would explain why it was $sec(x^2)tan(x^2)$, but how did he/she get $4x^2$ when the remaining was $(2)(2x) = 4x$ ?

10. Originally Posted by StarlitxSunshine
Oh..

Umm...But that would explain why it was $sec(x^2)tan(x^2)$, but how did he/she get $4x^2$ when the remaining was $(2)(2x) = 4x$ ?

You are correct there was a typo on my part.

It has been fixed above.

11. Ah, ohkay :3

But now, there are two different answers

$4xsec^2(x^2)tan(x^2)$ I understand how its done with the product & chain rule because you change the function to $(sec x^2)(sec x^2)$

But TheEmptySet has
$4xsec(x^2)tan(x^2)$ through substituting the parts of the chain rule. So shouldn't the answers be the same, because its essentially the same method ?

12. Originally Posted by StarlitxSunshine
Ah, ohkay :3

But now, there are two different answers

$4xsec^2(x^2)tan(x^2)$ I understand how its done with the product & chain rule because you change the function to $(sec x^2)(sec x^2)$

But TheEmptySet has
$4xsec(x^2)tan(x^2)$ through substituting the parts of the chain rule. So shouldn't the answers be the same, because its essentially the same method ?

Originally Posted by TheEmptySet
The function $y=\sec^2(x^2)$ if the composition of 3 functions. This means you need to use the chain rule.

This can be seen as

$y=s^2 \implies \frac{dy}{ds}=2s$
$s=sec(t) \implies \frac{ds}{dt}=\sec(t)\tan(t)$
$t=x^2 \implies \frac{dt}{dx}=2x$

So the chain rule tells us the derivative is

$\frac{dy}{dx}=\frac{dy}{ds}\frac{ds}{dt}\frac{dt}{ dx}=2s(\sec(t)\tan(t))(2x)$

Now just sub in from above to get

$\frac{dy}{dx}=2\sec(t)\tan(t)(2x)=4x\sec(x^2)\tan( x^2)$
he forgot the s the formula is like this

$\frac{dy}{dx}=2s\sec(t)\tan(t)(2x)=2(2x)\sec(x^2)\ sec(x^2)\tan(x^2)$[/quote]

......

13. oo0ohkay !

Thank you so much. After all of that, I should be able to do the question on my own without looking at your working, too !

Sorry for all the trouble >_<