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Math Help - Differentiating problems.

  1. #1
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    Differentiating problems.

    Okay, guys, I have two questions this time:

    1. If f(x) = {({\sqrt{{4x}^{2 } - 3 })^{\frac{-1}{2} then how does it become f(x) = ({x}^{2} -3)^{\frac{1}{2}} ? Shouldn't it become 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 f(x) = (cos{x}^{2}) f'(x) = (-2sinx) or (-sin{x}^{2})  (2x)? Is the {x}^{2} a separate term and thus differentiated separately? If that is the case, would I use the same convention to deal with f(x) = {sin}^{2}x?

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

    2. I can't tell if your original function is \displaystyle f(x) = \cos^2{x} or \displaystyle f(x) = \cos{(x^2)}. Please use brackets where they're needed to avoid ambiguity.
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  3. #3
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    For 1. the book just simply says "rewrite the function as (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 cos(x^{2}), so would it be differentiated as two terms or did they use the product rule to differentiate? Also, how would differentiating cos^{2}x and even (cosx^{2}) be different?
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    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 u=x^2 then \dfrac{d}{du} \cos(u)  \times \dfrac{d}{dx} x^2

    (but remember to back substitute)
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    Quote Originally Posted by e^(i*pi) View Post
    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 u=x^2 then \dfrac{d}{du} \cos(u)  \times \dfrac{d}{dx} x^2

    (but remember to back substitute)
    1. The original function is 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 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 (cosx^{2}) and cos^{2}x? For the former do we use the chain rule and thus differentiate as f'(x) = (-2sinx)(1)(cosx{^2})^{0} = > (-2sinx)? And for f(x) = (cos^{2}x) would it become
    f'(x) = (-sin^{2}x)(1)(cos^{2})^{0} = > (-sin^{2}x)?

    Thanks in advance for all the help.
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    Quote Originally Posted by Pupil View Post
    1. The original function is 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: f(x) = (4x^2-3)^{-1/4}

    Hereafter you can use the chain rule to give -\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 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 f(x) = x^2 \cos(x) then you'd need the product rule.

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

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

    Thanks in advance for all the help.
    It may help you visualise \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 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 -\sin(x)

    Multiplied this is -2\cos(x)\sin(x). If you know your double angle identities you may also write is as -\sin(2x)
    Last edited by e^(i*pi); May 27th 2011 at 02:50 PM. Reason: Adding missing "-" sign, see post below
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    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 -2x(4x^{2} - 3)^{\frac{-5}{4}} Instead of 2x(4x^{2} - 3)^{\frac{-5}{4}} ?
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    Quote Originally Posted by Pupil View Post
    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 -2x(4x^{2} - 3)^{\frac{-5}{4}} Instead of 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 (a^b)^c = a^{bc}
    Last edited by e^(i*pi); May 27th 2011 at 03:00 PM. Reason: see post
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