1. ## Differentiating problems.

Okay, guys, I have two questions this time:

1. If $\displaystyle f(x) = {({\sqrt{{4x}^{2 } - 3 })^{\frac{-1}{2}$ then how does it become $\displaystyle f(x) = ({x}^{2} -3)^{\frac{1}{2}}$? Shouldn't it become $\displaystyle f(x) =((4x-3)^{\frac{1}{2}})^{\frac{-1}{2}}$? My book did the in one step, before differentiating. The solution really threw me off.

2. Is the derivative of $\displaystyle f(x) = (cos{x}^{2})$ $\displaystyle f'(x) = (-2sinx)$ or $\displaystyle (-sin{x}^{2}) · (2x)$? Is the $\displaystyle {x}^{2}$ a separate term and thus differentiated separately? If that is the case, would I use the same convention to deal with $\displaystyle f(x) = {sin}^{2}x$?

Thanks in advance guys for all the help.

2. 1. I suggest you post the entire question, as there is no way I can see atm how $\displaystyle \displaystyle (\sqrt{4x^2 - 3})^{-\frac{1}{2}}$ can become $\displaystyle \displaystyle (x^2 - 3)^{\frac{1}{2}}$ without the context...

2. I can't tell if your original function is $\displaystyle \displaystyle f(x) = \cos^2{x}$ or $\displaystyle \displaystyle f(x) = \cos{(x^2)}$. Please use brackets where they're needed to avoid ambiguity.

3. For 1. the book just simply says "rewrite the function as $\displaystyle (x^{2}-3)^{\frac{1}{2}}$ before they begin to differentiate. There was no step in between to explain it, which threw me off.

For 2. I was referring to $\displaystyle cos(x^{2})$, so would it be differentiated as two terms or did they use the product rule to differentiate? Also, how would differentiating $\displaystyle cos^{2}x$ and even $\displaystyle (cosx^{2})$ be different?

4. Both of your questions appear to be about applications of the chain rule

1. What is the original function to be differentiated?

2. Normal chain rule. If you let $\displaystyle u=x^2$ then $\displaystyle \dfrac{d}{du} \cos(u) \times \dfrac{d}{dx} x^2$

(but remember to back substitute)

5. Originally Posted by e^(i*pi)
Both of your questions appear to be about applications of the chain rule

1. What is the original function to be differentiated?

2. Normal chain rule. If you let $\displaystyle u=x^2$ then $\displaystyle \dfrac{d}{du} \cos(u) \times \dfrac{d}{dx} x^2$

(but remember to back substitute)
1. The original function is $\displaystyle f(x) = {({\sqrt{{4x}^{2 } - 3 })^{\frac{-1}{2}$, so would you say the book probably made an error and differentiated for a different function?

As for 2, to apply the chain rule don't both terms need to be in the brackets? So, wouldn't it be the product rule to differentiate $\displaystyle f(x) = cos(x^{2})$? I am not saying you are wrong, but I seem to have the most trouble with this type of function.

Also, how would you differentiate $\displaystyle (cosx^{2})$ and $\displaystyle cos^{2}x$? For the former do we use the chain rule and thus differentiate as $\displaystyle f'(x) = (-2sinx)(1)(cosx{^2})^{0} = > (-2sinx)$? And for $\displaystyle f(x) = (cos^{2}x)$ would it become
$\displaystyle f'(x) = (-sin^{2}x)(1)(cos^{2})^{0} = > (-sin^{2}x)$?

Thanks in advance for all the help.

6. Originally Posted by Pupil
1. The original function is $\displaystyle f(x) = {({\sqrt{{4x}^{2 } - 3 })^{\frac{-1}{2}$, so would you say the book probably made an error and differentiated for a different function?
I'm not sure, I was unsure before as to what the original function was.

Do you understand why we can rewrite: $\displaystyle f(x) = (4x^2-3)^{-1/4}$

Hereafter you can use the chain rule to give $\displaystyle -\dfrac{1}{4}(4x^2-3)^{-5/4} \cdot 8x = -2x(4x^2-3)^{-5/4} = -\dfrac{2x}{\sqrt[4]{(4x^2-3)^5}}$.

As for 2, to apply the chain rule don't both terms need to be in the brackets? So, wouldn't it be the product rule to differentiate $\displaystyle f(x) = cos(x^{2})$? I am not saying you are wrong, but I seem to have the most trouble with this type of function.
No, you have a function (x^2) within another function cos(x^2) so the chain rule applies. The product rule applies if you have two functions multiplied together. For example if $\displaystyle f(x) = x^2 \cos(x)$ then you'd need the product rule.

Also, how would you differentiate $\displaystyle (cosx^{2})$
The same as above, using the chain rule. If possible avoid this notation as it's unclear.

and $\displaystyle cos^{2}x$? For the former do we use the chain rule and thus differentiate as $\displaystyle f'(x) = (-2sinx)(1)(cosx{^2})^{0} = > (-2sinx)$? And for $\displaystyle f(x) = (cos^{2}x)$ would it become
$\displaystyle f'(x) = (-sin^{2}x)(1)(cos^{2})^{0} = > (-sin^{2}x)$?

Thanks in advance for all the help.
It may help you visualise $\displaystyle \cos^2(x) = (\cos(x))^2$. You can then use the chain rule. I'm not sure where your 0 power comes from though.

If you use the power rule as normal (ignoring the cos(x) for now and leaving it there) we get $\displaystyle 2\cos(x)$. However, we're not done since we need to multiply by the derivative of cos(x) [which we ignored previously] which is $\displaystyle -\sin(x)$

Multiplied this is $\displaystyle -2\cos(x)\sin(x)$. If you know your double angle identities you may also write is as $\displaystyle -\sin(2x)$

7. Thanks, e^(i*pi). That cleared up everything for me. One last question: whenever you have a power inside a bracket and outside the bracket, you can multiply them out like you just did so it can be easier to simplify, correct? Also, shouldn't it it become $\displaystyle -2x(4x^{2} - 3)^{\frac{-5}{4}}$ Instead of $\displaystyle 2x(4x^{2} - 3)^{\frac{-5}{4}}$ ?

8. Originally Posted by Pupil
Thanks, e^(i*pi). That cleared up everything for me. One last question: whenever you have a power inside a bracket and outside the bracket, you can multiply them out like you just did so it can be easier to simplify, correct? Also, shouldn't it it become $\displaystyle -2x(4x^{2} - 3)^{\frac{-5}{4}}$ Instead of $\displaystyle 2x(4x^{2} - 3)^{\frac{-5}{4}}$ ?
I was just testing you, honest

Yes, you're right, it should be -2x out front, I'll just edit my previous post to that effect

edit: Yes you can do that with powers, it's using the law of exponents that says $\displaystyle (a^b)^c = a^{bc}$