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    https://gyazo.com/42efc6c5f2c30e833f66b33b65399210

    y^e^(x) = 2^x . cos(x)

    my method so far:

    e^(x) . ln(y) = xln(2) + ln(cosx)

    ^ product rule for LHS & RHS you can differentiaite indiivudually giving

    RHS' = ln(2) + 1/cosx . -sinx
    RHS' = ln(2) - tan(x)

    LHS I can just use product rule with e^(x) . ln(y) but implicitly right?

    https://gyazo.com/786fec9a25c4a5609521c792720d7d31

    root( x - lny ) = sin (xy)

    my method so far:

    x - lny = sin^2 (xy) ----> from here I thought it can be rewritten as
    x - lny = [sin(xy)]^2 and on the RHS I can just use the chain rule

    RHS' = 2[sin(xy)] . cos(xy) .....----> but then I have to use the product rule to differentiate the inside bracket of (x.y) right? so I'd have the entire RHS differentiated correct?

    The LHS I can differentiate individually fine.

    I'd appreciate if someone can guide/tell me if my methods are correct or how to go about solving them correctly if I made a mistake etc.
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    (Original post by XxKingSniprxX)
    https://gyazo.com/42efc6c5f2c30e833f66b33b65399210

    y^e^(x) = 2^x . cos(x)

    my method so far:

    e^(x) . ln(y) = xln(2) + ln(cosx)

    ^ product rule for LHS & RHS you can differentiaite indiivudually giving

    RHS' = ln(2) + 1/cosx . -sinx
    RHS' = ln(2) - tan(x)

    LHS I can just use product rule with e^(x) . ln(y) but implicitly right?

    https://gyazo.com/786fec9a25c4a5609521c792720d7d31

    Looks good so far. Can you continue and post your attempt if you get stuck?

    root( x - lny ) = sin (xy)

    my method so far:

    x - lny = sin^2 (xy) ----> from here I thought it can be rewritten as
    x - lny = [sin(xy)]^2 and on the RHS I can just use the chain rule

    RHS' = 2[sin(xy)] . cos(xy) .....----> but then I have to use the product rule to differentiate the inside bracket of (x.y) right? so I'd have the entire RHS differentiated correct?

    The LHS I can differentiate individually fine.

    I'd appreciate if someone can guide/tell me if my methods are correct or how to go about solving them correctly if I made a mistake etc.
    f = \sin\left(xy\right)^2

    f' = 2\sin\left(xy\right)\times \left(\sin\left(xy\right) \right)'

    So now you need to consider the derivative of \sin\left(xy\right). You can treat this as a separate question and use the chain rule. Can you try that?

    For complicated chain rule, it's useful to present your working like I've done with \left(\sin\left(xy\right) \right)' then work out what \left(\sin\left(xy\right) \right)' is on a separate line then put it back into the original derivative.
 
 
 
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