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Thread: Deriving exponential/ln functions

  1. #1
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    Deriving exponential/ln functions

    The question is to differentiate the following expression:

    Deriving exponential/ln functions-q3.jpg

    My attempt was as follows:

    Deriving exponential/ln functions-q3w1.jpg
    Deriving exponential/ln functions-q3w2.jpg
    Deriving exponential/ln functions-q3w3.jpg

    However the answer is:

    Deriving exponential/ln functions-q3a.jpg

    I was wondering where I was going wrong?

    Thanks in advance!
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  2. #2
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    Re: Deriving exponential/ln functions

    Quote Originally Posted by ImPoob View Post
    The question is to differentiate the following expression:
    Click image for larger version. 

Name:	q3.JPG 
Views:	11 
Size:	8.7 KB 
ID:	39265
    What you have written is best expressed using notation:
    $y=\exp(2\log(sin(x))$, that is using web-convenient notation: $e^t=\exp(t)~\&~ln(x)=\log(x)$
    Because $\exp(\log(T))=T$ what you posted is $y=\exp(2\log(\sin(x)))=\exp(\log(\sin^2(x)))=\sin ^2(x)$.
    Thus if $y=\sin^2(x)$ then $y'=2(\sin(x))(\cos(x))$
    Can you finish ?
    Last edited by Plato; Feb 10th 2019 at 06:27 PM.
    Thanks from topsquark
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  3. #3
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    Re: Deriving exponential/ln functions

    Ahh I see thank you. You would just use the identity to rewrite 2sin(x)cos(x) = sin(2x). Could you take a moment to explain why we cannot derive it as we would normally with exp functions. e.g. we would derive y = e^3x as dy/dx = e^3x . 3

    Thanks a bunch again
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  4. #4
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    Re: Deriving exponential/ln functions

    You can, but it's easier to simplify the expression first.
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  5. #5
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    Re: Deriving exponential/ln functions

    In your attempt you stopped at $\displaystyle \frac{dy}{dx}= 2cot(x)e^{2 ln(sin(x))}= \frac{2cos(x)}{sin(x)}e^{2 ln(sin(x))}$.

    Now, as Plato suggested, use $\displaystyle e^{2 ln(sin(x))}= e^{ln(sin^2(x))}= sin^2(x)$ to reduce that to $\displaystyle \frac{dy}{dx}= \frac{2 cos(x)}{sin(x)}sin^2(x)= 2 sin(x)cos(x)$.
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