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    a curve y=f(x) is given implicitly by the equation
    y3 +x3y+tanxy2-x4=4

    Find then gradient at x=0

    Thanks for the help
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    (Original post by achu2k11)
    a curve y=f(x) is given implicitly by the equation
    y3 +x3y+tanxy2-x4=4

    Find then gradient at x=0

    Thanks for the help
    Do you know how to differentiate implicitly?
    What is your particular problem?
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    Differentiate each term with respect to x.

    For example \displaystyle \frac{d}{dx}(y^3)=3y^2 \frac{dy}{dx}.

    Once you've done that, solve for dy/dx.

    Find the value of y when x=0. Put those x and y values in your expression for dy/dx.

    Does that jog your memory?
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    (Original post by joostan)
    Do you know how to differentiate implicitly?
    What is your particular problem?
    Yeah, Can you check if this is right

    dy/dx (3x2+y2sec2(x)-4x3)=3y2+2ytan(x)
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    (Original post by achu2k11)
    Yeah, Can you check if this is right

    dy/dx (3x2+y2sec2(x)-4x3)=3y2+2ytan(x)
    I disagree with this. . .
    You also need to use the product rule
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    (Original post by achu2k11)
    Yeah, Can you check if this is right

    dy/dx (3x2+y2sec2(x)-4x3)=3y2+2ytan(x)
    You are sort of getting there but how are you differentiating exactly?

    how did you differentiate x3y and and tan(x)y2?
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    (Original post by hoodboilu4)
    You are sort of getting there but how are you differentiating exactly?

    how did you differentiate x3y and and tan(x)y2?
    Isn't using product rule of differentiation
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    (Original post by achu2k11)
    Isn't using product rule of differentiation
    Yes so what were you get when you differentiated those two?
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    (Original post by hoodboilu4)
    Yes so what were you get when you differentiated those two?
    Is it:

    3x2dy/dx y+x3

    and

    y2sec(x)dy/dx+2ytan(x)
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    (Original post by achu2k11)
    Is it:

    3x2dy/dx y+x3

    and

    y2sec(x)dy/dx+2ytan(x)
    You seem to be getting confused with your \frac{dy}{dx}s
    \dfrac{d}{dx}(x^ny^m)=nx^{n-1}y^m + x^nmy^{m-1}\frac{dy}{dx}
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    (Original post by achu2k11)
    Is it:

    3x2dy/dx y+x3

    and

    y2sec(x)dy/dx+2ytan(x)
    It's supposed to be 3x2y + x3 dy/dx

    and y2sec(x) + 2ytan(x)dy/dx
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    (Original post by hoodboilu4)
    It's supposed to be 3x2y + x3 dy/dx

    and y2sec(x) + 2ytan(x)dy/dx
    You mean y^2\sec^2(x) + 2y\tan(x)\frac{dy}{dx}
    You're missing a squared
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    (Original post by joostan)
    You mean y^2\sec^2(x) + 2y\tan(x)\frac{dy}{dx}
    You're missing a squared
    I am indeed :P
 
 
 
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