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Math Help - Calculus Questions

  1. #1
    Newbie midsummer's Avatar
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    Calculus Questions

    Hi I need help with these questions:

    1.Write the composite function in the form f(g(x)). then find the derivative dy/dx. y=sin square root x.

    2. Find the first and second derivatives of the function y=4x/ square root (X+1)

    3. find dy/dx by implicit differentiation
    square root (x+y)=1 +x^2 y^2
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  2. #2
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    1) f(x)=sin x, g(x)=\sqrt{x}
    by chain rule: dy/dx=f'(g(x))g'(x). Try by yourself first!
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  3. #3
    Newbie midsummer's Avatar
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    I got cos square root x/ 2 square root x. Not sure if this is correct. :/
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  4. #4
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    Quote Originally Posted by midsummer View Post
    I got cos square root x/ 2 square root x. Not sure if this is correct. :/
    You need to be confidence on yourself


    For number 2, use quotient rule.

    For number 3, you have
    \sqrt{x+y}=1+x^2y^2

    I'll do the LHS and you do the RHS
    <br />
\frac{d}{dX}(x+y)^{1/2}=\frac{1}{2} (x+y)^{-1/2}\cdot \frac{d}{dx} (x+y)=\frac{1}{2} (x+y)^{-1/2} (1+\frac{dy}{dx})

    After you do the RHS, then solve the equation for dy/dx
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  5. #5
    Newbie midsummer's Avatar
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    So is my answer to the first question correct?

    As for number 2. I got y= 2x+1/ 2(x+1)^1/2. Not sure about that :/

    As for number 3. I am still lost :/

    It's not that I lack confidance, I am trying to get better with practice and it's hard when I get stuck on questions. I really appreciate the help
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by midsummer View Post
    3. find dy/dx by implicit differentiation
    square root (x+y)=1 +x^2 y^2
    In implicit differentiation, you treat x and y as functions of x.

    So when you differentiate \sqrt{x+y}=1+x^2y^2, you need to apply chain rule to \sqrt{x+y} and product rule to x^2y^2

    I'll show the initial first step, and then I'll let you simplify this.

    \frac{\,d}{\,dx}\left[\sqrt{x+y}\right]=\frac{\,d}{\,dx}\left[1+x^2y^2\right]\implies \tfrac{1}{2}(x+y)^{-\frac{1}{2}}\cdot\left(1+\frac{\,dy}{\,dx}\right)=  2xy^2+2x^2y\frac{\,dy}{\,dx}\implies\dots

    From here, simplify a little bit, but the important part is to group the terms that contain \frac{\,dy}{\,dx} together. Then we can easily solve for \frac{\,dy}{\,dx}.

    Can you take it from here?

    --Chris
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