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Math Help - Fun with Implicit Derivatives and lns

  1. #1
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    Fun with Implicit Derivatives and lns

    Hello. I have two super fun problems for you guys to try and solve. I tried them both but my answers were not right.

    2e^{4-x^2}-4ln(x^3+5)

    Answer is supposed to be -4xe^{4-x^2} - \frac{12x^2}{x^3+5}
    and

    2y^3+x^2y^2=5x-y-1

    This one is suposed to be \frac{1}{3}

    If you could be so kinds as to show me your steps I would greatly appreciate it. I think i might be messing up my simplification on the first one i tried it 3 times and i got the same answer all 3 times but its wrong according to the key and I doubt my physics prof is wrong. The 2nd one makes me want to cut myself
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  2. #2
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    2y^{3}+x^{2}y^{2}-5x+y+1=0

    One common mistake with implicit diff is not using the product rule.

    6y^{2}y'+\left(\underbrace{2x^{2}yy'+2xy^{2}}_{\te  xt{product rule }}\right)-5+y'=0

    y'=\frac{5-2xy^{2}}{6y^{2}+2x^{2}y+1}

    Where is the 1/3 coming from?. Are there values given for x and y?
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  3. #3
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    2e^{4-x^2}-4ln(x^3+5)
    I see nothing particularly "fun" or very difficult about this. The derivative of e^u with respect to u is e^u and using the chain rule the derivative of 2e^{4- x^2} is 2(e^{4- x^2})(-2x)= -4xe^{4-x^2}. The derivative of ln(u) with respect to u is 1/u so the derivative of -4ln(x^3+ 5) is (-4)(\frac{1}{x^3+5})(3x^2)= \frac{-12x^2}{x^3+ 5}. Adding those gives what you say the answer is "supposed" to be.
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