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Math Help - Derivative problems

  1. #1
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    Derivative problems

    I need help with the following derivative probelms. They alll need the chain rule.

    1) y = x/sqrt(x^4+2)

    2) g(t) = 5 cos pi(t)

    3) h(x) = ln(2x^3+3)

    Thanks for any and all help.
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  2. #2
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    Don't be put off by the chain rule. Sometimes it is presented in a complicated manner, but it's really the derivative of the outside times the derivative of the inside.

    Here's the first one. Let's change it to a product and use the product rule.
    It's a good idea to do that when possible. The product rule is easier to work with than the quotient rule in most cases.

    x(x^{4}+2)^{\frac{-1}{2}}

    x\overbrace{(\frac{-1}{2})(x^{4}+2)^{\frac{-3}{2}}}^{\text{derivative of outside}}\underbrace{(4x^{3})}_{\text{derivative of inside}}+(x^{4}+2)^{\frac{-1}{2}}(1)

    Now, hammer it into shape with some algebra:

    \frac{-2x^{4}}{(x^{4}+2)^{\frac{3}{2}}}+\frac{1}{(x^{4}+2  )^{\frac{1}{2}}}

    \frac{-2x^{4}}{(x^{4}+2)^{\frac{3}{2}}}+\frac{1}{(x^{4}+2  )^{\frac{1}{2}}}\cdot\frac{(x^{4}+2)}{(x^{4}+2)}

    \boxed{\frac{2-x^{4}}{(x^{4}+2)^{\frac{3}{2}}}}
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  3. #3
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    you can use quotient or product rule for 1st one.
    you'll get -2x^4 /(2+x^4)^1.5 + 1/(2+x^4)^0.5 if you do it right.

    2. just differentiate cos then differentiate the inside...
    g'(t) = -5π*sin (πt)

    3. if ln [f(x)], then d/dx ln [f(x)] = f'(x)/f(x)
    h'(x) = 6x/[2x+3]
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  4. #4
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    Thanks guys, the last one was so easy!! Anyway I still dont understand the second. Thanks.
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  5. #5
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    if you have a function cos[f(x)], then d/dx cos[f(x)] = -f'(x)*sin[f(x)]
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  6. #6
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    Im still confused on the second one. By the way pi = the symbol pi or 3.14

    Thanks.
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  7. #7
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    pi is just a constant.
    d/dx cos[f(x)] = -f'(x)*sin[f(x)]
    apply this to your question
    d/dx 5cos(πt)

    5 at the front is just a constant, so just leave it.
    formula says (once you multiply both sides by 5) => 5*d/dx cos[f(x)] = -5*f'(x)*sin[f(x)] your f(x) in this case is πt, so f'(x) = π.
    so sub it in...
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