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Thread: Differentiating...

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    Differentiating...

    Okay, so I'm finding the derivative of y = (5x - 1)^2 (3x + 7)^3

    I've gotten it to y' = (5x - 1)^2 times 9(3x + 7)^2 + 10(5x-1) times (3x + 7)^3


    My question is: do I combine all of these terms together or do you typically just leave it in this form?
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    Quote Originally Posted by bobsanchez View Post
    Okay, so I'm finding the derivative of y = (5x - 1)^2 (3x + 7)^3

    I've gotten it to y' = (5x - 1)^2 times 9(3x + 7)^2 + 10(5x-1) times (3x + 7)^3


    My question is: do I combine all of these terms together or do you typically just leave it in this form?
    It may not be the most compact way to express the derivative,
    but if you substitute a value of x into that, you will get the curve gradient at that x.

    You can simplify further...


    10(5x-1)(3x+7)^3+9(3x+7)^2(5x-1)^2

    has (3x+7)^2(5x-1) common


    (3x+7)^2(5x-1)\left(10[3x+7]+9[5x-1]\right)=(3x+7)^2(5x-1)(75x+61)

    is fairly compact
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    Another similar "procedural" type question: is it necessary to put in terms of y in order to differentiate? For instance...

    Differentiate 17cos(x)sin(y) = 12

    Would I have to put in terms of y and, if so, how would I do that? If I can simply differentiate straightforwardly, I think I can do it.
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    Quote Originally Posted by bobsanchez View Post
    Another similar "procedural" type question: is it necessary to put in terms of y in order to differentiate? For instance...

    Differentiate 17cos(x)sin(y) = 12

    Would I have to put in terms of y and, if so, how would I do that? If I can simply differentiate straightforwardly, I think I can do it.
    I recommend you use implicit differentiation.
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    what happens with the 12?
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    Quote Originally Posted by bobsanchez View Post
    what happens with the 12?
    what's the derivative of a constant?
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    Okay, so 17cos(x)cos(y) - 17sin(x)sin(y) = 0?
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    Quote Originally Posted by bobsanchez View Post
    Okay, so 17cos(x)cos(y) - 17sin(x)sin(y) = 0?
    You've lost the derivative!

    17cosxsiny=12

    cosxsiny=\frac{12}{17}

    If differentiating wrt x

    siny(-sinx)+cosx\frac{dy}{dx}\frac{d}{dy}siny=0

    using the chain rule
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    \displaystyle \frac{d}{dx}\left[17\cos{x}\sin{y} = 12\right]<br />

    \displaystyle \frac{d}{dx}\left[\cos{x}\sin{y} = \frac{12}{17}\right]<br />

    \displaystyle<br />
\cos{x}\cos{y} \cdot \frac{dy}{dx} + \sin{y}(-\sin{x}) = 0

    \displaystyle<br />
\frac{dy}{dx} = \frac{\sin{x}\sin{y}}{\cos{x}\cos{y}} = \tan{x}\tan{y}
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