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Thread: 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.

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

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

    $\displaystyle 2y^3+x^2y^2=5x-y-1$

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

    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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    $\displaystyle 2y^{3}+x^{2}y^{2}-5x+y+1=0$

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

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

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

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