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    Okay so this is my first set of hw for advanced maths and tbh I'm struggling and it's level c questions!!
    If anyone could help with any of 1m, 4, 5, 8b and all of 9 that would be great!!

    The answers are
    1m. -3/ 1+sin3x
    4 and 5 is proof
    8b. 25
    9 is all proof

    Working would be really helpful as well! And I'm not a complete fool as I managed to complete my other hw fine but I just can't do this differentiation that great yet but practice makes perfect I guess! Thanks in advance
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    (Original post by grumpylamb)
    ...
    Looking at 1m to start, what's the problem? If you've done the others, I presume you're fairly competent with the product/quotient rules and just having problems with the algebra.

    Post some working.
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    (Original post by grumpylamb)
    Okay so this is my first set of hw for advanced maths and tbh I'm struggling and it's level c questions!!
    If anyone could help with any of 1m, 4, 5, 8b and all of 9 that would be great!!

    The answers are
    1m. -3/ 1+sin3x
    4 and 5 is proof
    8b. 25
    9 is all proof

    Working would be really helpful as well! And I'm not a complete fool as I managed to complete my other hw fine but I just can't do this differentiation that great yet but practice makes perfect I guess! Thanks in advance
    Posted from TSR Mobile
    for 1m
    Use the quotient rule and simplify
    \displaystyle \frac{d}{dx}\left (\frac{f(x)}{g(x)}\right )=\frac{\frac{df(x)}{dx}\cdot g(x)-\frac{dg(x)}{dx}\cdot f(x)}{\left [g(x)\right]^2}
    Here
    \displaystyle f(x)=1+\sin 3x
    \displaystyle g(x)=\cos 3x

    for 4
    Diffeerentiate y with respect to theta using the quotient rule -> y'=\frac{dx}{dx}
    Differentiate one more y' -> y" =\frac{d^2y}{dx^2}
    Substitute these in the equation.

    for 5
    Differentiate y twice using the product rule then the quotient rule, and use some trigonoimetric identities
    \displaystyle \frac{dy}{dx}=e^x\cdot \tan x +e^x\cdot \frac{1}{\cos^2 x}=e^x\cdot \left (\tan x+\frac{1}{\cos^2 x}\right )
    \displaystyle \frac{d^2y}{dx^2}=e^x\cdot \left (\tan x+\frac{1}{\cos^2 x}\right )+e^x\cdot \left (\frac{1}{\cos^2 x}+\frac{2\cdot \sin x \cdot \cos x}{\cos^4 x}\right )
    Taking out e^x and using trigonometric identities
    \displaystyle e^x \cdot \left ( \tan x + 1+\tan^2 x +1+\tan^2 x+2\tan x \cdot (1+\tan^2 x)\right )
    From this you will get the answer

    8b)
    \displaystyle A^2=25x^2-\frac{1}{4}x^4
    Differentiate both side with respect to x
    on the LHS and RHS
    2\cdot A \cdot \frac{dA}{dx} =50x-x^3
    divide by 2A and you will get A'-
    \displaystyle A'=\frac{50x-x^3}{50x^2-\frac{1}{2}x^4}
    A has maximujm when A'=0

    for 9
    THe perimeter
    \displaystyle 5=2r+2h+r\pi
    solve for h

    THe area
    \displaystyle A=\frac{1}{2}r^2 \cdot \pi+2r\cdot h
    Sub h the terms above
    For maximum
    \displaystyle \frac{dA}{dr}=0
 
 
 
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