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Math Help - [SOLVED] Differentiation

  1. #1
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    Exclamation [SOLVED] Differentiation

    Differentiate with respect to x: x^3siny - y^3cos = x^3

    I've never done this sort of differentiation before. Could someone please show me how it's done?
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Differentiate with respect to x: x^3siny - y^3cos = x^3

    I've never done this sort of differentiation before. Could someone please show me how it's done?
    some good examples and explanation here ...

    http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html

    go learn, try your problem, then receive feedback from your attempt.
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  3. #3
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    Hello, xwrathbringerx!

    You've never done Implicit Differentiation?
    Then who assigned this problem to you?

    Besides, you left out part of the problem . . .


    Differentiate with respect to x\!:\;\;x^3\sin y - y^3\cos {\color{red}?} \:=\: x^3

    I'll assume the problem is: . x^3\sin y - y^3\cos{\color{red}x} \:=\:x^3


    We have: . x^3\cos y\!\cdot\!\frac{dy}{dx} + 3x^2\sin y - y^3(\text{-}\sin x) - 3y^2\cos x\!\cdot\!\frac{dy}{dx} \:=\:3x^2

    . . . . . . x^3\cos y\!\cdot\!\frac{dy}{dx} - 3y^2\cos x\!\cdot\!\frac{dy}{dx} \:=\:3x^2-3x^2\sin y - y^3\sin x

    Factor: . \left(x^3\cos y - 3y^2\cos x\right)\,\frac{dy}{dx} \:=\:3x^2-3x^2\sin y - y^3\sin x


    Therefore: . \frac{dy}{dx} \;=\;\frac{3x^2-3x^2\sin y - y^3\sin x}{x^3\cos y - 3y^2\cos x}

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