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    Y= 7x(Cosx)^(x/2)

    Q. FIND DY/DX

    1- Ln both sides

    Working on the RHS first

    Ln(7x(Cosx)^(x/2))

    2. Then "Expand the bracket" using logaritmic rules=

    Ln(7x) +Ln(Cosx)^(x/2)

    3. Now, Bring down the power in the second term to the front

    Ln(7x) + (x/2)Ln Cosx

    4. Differenriate as usual with respect to x

    First term

    Ln(7x) = 1/(7x)

    For the Second term

    (X/2) LnCos X

    Use the product rule and the chain rule because there are 2 functions being multiplied and an inner and outer function.

    Using the product rule to differentiate (x/2)Ln Cosx

    U= x/2 V= lnCosx

    U'v +V'u=

    (X/2) (LnCosx')+ LnCosx(X/2')

    Differentiating LnCos x requires the Chain rule

    = (1/Cosx)* -Sinx
    Since sinx/cosx = Tan x, -(sinx/cosx) = -tanx

    second term (of the product rule) - derivative of x/2= 1/2


    So together we have:

    (X/2)*-tanx+LnCosx(1/2)

    Simplifying:

    (-tanx+lncosx)/2

    4. Working on the LHS now

    Lny=( 1/y)dy/dx

    Therefore the whole thing is:

    (1/Y)dy/dx= -Tanx+Lncosx/2

    Multiplying both sides by y we get:

    Dy/dx= Y *( -Tanx+Lncosx/2)

    Substituting the value of Y from the equation

    Recall: Y= 7x(cosx)^(x/2)

    Therefore dy/dx=

    7x(cosx)^(x/2)* ((-Tanx+Lncosx)/2)



    Please can anyone check this for me, I would be ever so grateful!! 😊😊😊
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    (Original post by FloralEssence)
    ...
    No that's not right. Also PLEASE use LaTex or clearer representation, there's a lot of ambiguity there.

    \frac{d}{dx}(\ln \lvert 7x \lvert )=\frac{1}{x}\not= \frac{1}{7x}

    and when you say "the whole thing is..." you are missing the derivative I corrected above.

    \frac{d}{dx}(\frac{x}{2} \ln\lvert \cos x \lvert ) is correct though.

    ALSO please do not say \ln(7x) = \frac{1}{7x} as it doesn't make sense when you're actually differentiating it...

    (the more i look through this the more errors i seem to find)
 
 
 
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