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anonymous

  • 5 years ago

fine i am puttin ma questions here

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  1. anonymous
    • 5 years ago
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    1st of all in the question \\[y = sinx ^{x ^{2}}\]

  2. anonymous
    • 5 years ago
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    here i could not understand i got till \[d/dx(\log y) = d/dx(x^2 \log (\sin x))\] but am not able to proceed further

  3. anonymous
    • 5 years ago
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    The chain rule states that, \[d/dx(f(y)) = f'(y) dy/dx\]

  4. anonymous
    • 5 years ago
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    ok lets go slowly. your problem is that the variable is in the exponent. you get it out of the exponent by taking the log

  5. anonymous
    • 5 years ago
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    yups

  6. anonymous
    • 5 years ago
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    so first step is to take the log

  7. anonymous
    • 5 years ago
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    now d/dx(logy) = 1/y dy/dx? Did you get this?

  8. anonymous
    • 5 years ago
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    did that as u saw!!!

  9. anonymous
    • 5 years ago
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    no amogh i did not get this !!!!!

  10. anonymous
    • 5 years ago
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    \[ln(sin(x))^{x^2}=x^2ln(sin(x))\] by the properties of logs

  11. anonymous
    • 5 years ago
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    so far so good?

  12. anonymous
    • 5 years ago
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    that's wat i did till here i got !!!!!!

  13. anonymous
    • 5 years ago
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    see above !!!!!

  14. anonymous
    • 5 years ago
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    satellite: I suggest you to start using a basic chain rule example as he stumbles only upon the chain part! good luck

  15. anonymous
    • 5 years ago
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    ok next step is to take the derivative. \[\frac{d}{dx}x^2ln(sin(x))\] and for this you need both the product rule and the chain rule.

  16. anonymous
    • 5 years ago
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    dude i did this already!!!!

  17. anonymous
    • 5 years ago
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    so now i apply chain rule????????

  18. anonymous
    • 5 years ago
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    as it is both a product, \[x^2\times ln(sin(x))\] and \[ln(sin(x))\] is a composite function. is the the log OF the sine of x

  19. anonymous
    • 5 years ago
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    wat do u mean by a composite function??????

  20. anonymous
    • 5 years ago
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    in english, the derivative of a composite function is the derivative of the outside function evaluated at the inside function, times the derivative of the inside function.

  21. anonymous
    • 5 years ago
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    i got confused!!!!!!

  22. anonymous
    • 5 years ago
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    composite means \[f(g(x))\] so if \[f(x)=ln(x)\] and \[g(x) = sin(x)\] then \[f(g(x))=ln(g(x))=ln(sin(x))\]

  23. anonymous
    • 5 years ago
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    wat happened to the outer x^2????

  24. anonymous
    • 5 years ago
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    when i diff (log y )i get i/y

  25. anonymous
    • 5 years ago
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    this is a composition of functions. another example. \[f(x)=\sqrt{x}\] \[g(x)=1-x^2\] \[f(g(x))=f(1-x^2)=\sqrt{1-x^2}\]

  26. anonymous
    • 5 years ago
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    oh no. the derivative of \[ln(x)=\frac{1}{x}\]

  27. anonymous
    • 5 years ago
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    I am saying the same for th LHS ie log y

  28. anonymous
    • 5 years ago
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    but the derivative of \[ln(g(x))=\frac{1}{g(x)}\times g'(x)\]

  29. anonymous
    • 5 years ago
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    so the derivative of \[ln(sin(x))=\frac{1}{sin(x)}\times cos(x)=\frac{cos(x)}{sin(x)}\]

  30. anonymous
    • 5 years ago
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    i am going slow and still wanting to take the derivative of \[x^2ln(sin(x))\] correctly. we will deal with the consequence of taking the log later.

  31. anonymous
    • 5 years ago
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    yes here is where i am stuck

  32. anonymous
    • 5 years ago
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    x^2 becomes 2x right???????

  33. anonymous
    • 5 years ago
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    right

  34. anonymous
    • 5 years ago
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    you have a product here. do you know the product rule?

  35. anonymous
    • 5 years ago
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    yay so we get in the end 2x.(ln(sin(x)))

  36. anonymous
    • 5 years ago
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    product rule says dat 1st take the 1st constant and diff the other and add the vis - a - vis to it!!!!!!

  37. anonymous
    • 5 years ago
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    nope sorry. again lets to slow

  38. anonymous
    • 5 years ago
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    CHEATING!!!!!!! ( cranky kid's voice)!!!!!!!!

  39. anonymous
    • 5 years ago
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    the derivative of \[f(x)g(x)\] is \[f'(x)g(x)+g'(x)g(x)\]

  40. anonymous
    • 5 years ago
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    well dat's wat i said in words just to avoid the typing!!!!!!!

  41. anonymous
    • 5 years ago
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    put \[f(x)=x^2\] \[f'(x)=2x\]

  42. anonymous
    • 5 years ago
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    yups i know f'(x)

  43. anonymous
    • 5 years ago
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    \[g(x)=ln(sin(x))\]

  44. anonymous
    • 5 years ago
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    \[g'(x)=\frac{cos(x)}{sin(x)}\]

  45. anonymous
    • 5 years ago
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    now dis is de place where i get stuck up y do we get dis????????????

  46. anonymous
    • 5 years ago
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    ahh that is the "chain rule"

  47. anonymous
    • 5 years ago
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    i say it in english. the derivative of log x is one over x

  48. anonymous
    • 5 years ago
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    how is this the chain rule i mean i just read it for algebra not for log or sin

  49. anonymous
    • 5 years ago
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    but the derivative of log of "something" is one over something times the derivative of " something"

  50. anonymous
    • 5 years ago
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    so by dat logic ln(sin(x) ) when derivatived we shud get 1/sin(x)

  51. anonymous
    • 5 years ago
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    right

  52. anonymous
    • 5 years ago
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    times the derivative of sine(x)

  53. anonymous
    • 5 years ago
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    which is...

  54. anonymous
    • 5 years ago
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    a rule?????????

  55. anonymous
    • 5 years ago
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    do you know the derivative of sin(x)?

  56. anonymous
    • 5 years ago
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    nope

  57. anonymous
    • 5 years ago
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    guess.

  58. anonymous
    • 5 years ago
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    i know dat u r a gud teacher and u r real patient thanx for dat!!!!!!

  59. anonymous
    • 5 years ago
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    is it cos(x)

  60. anonymous
    • 5 years ago
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    yes!

  61. anonymous
    • 5 years ago
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    so i repeat but the derivative of log of "something" is one over something times the derivative of " something"

  62. anonymous
    • 5 years ago
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    so we multiply the cos(x)

  63. anonymous
    • 5 years ago
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    ok ca we quickly repeat!!!!!!!!!!!!!

  64. anonymous
    • 5 years ago
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    so we get the derivative of \[ln(sin(x))\] is \[\frac{1}{sin(x)}\times cos(x)=\frac{cos(x)}{sin(x)}\]\]

  65. anonymous
    • 5 years ago
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    whew got it.

  66. anonymous
    • 5 years ago
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    so we get 1/y = 2x. cos(x)/sin(x) true???

  67. anonymous
    • 5 years ago
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    srsly whew my brain is alrdy stewd!!!!!

  68. anonymous
    • 5 years ago
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    now we put this all back in the product rule (FG)'=F'G+G'F \[F+x^2\] \[F'=2x\] \[G=ln(sin(x))\] \[G'=\frac{cos(x)}{sin(x)}\]

  69. anonymous
    • 5 years ago
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    i meant \[F=x^2\]

  70. anonymous
    • 5 years ago
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    yes go on!!!!! i am trying to get a full pic

  71. anonymous
    • 5 years ago
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    to get .... the derivative of \[x^2ln(sin(x))=2xln(sin(x))+x^2 \frac{cos(x)}{sin(x)}\]

  72. anonymous
    • 5 years ago
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    that is the product rule, although we used the chain rule to find the derivative of \[ln(sin(x))\]

  73. anonymous
    • 5 years ago
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    yes!!!!!!!

  74. anonymous
    • 5 years ago
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    whew. of course we are not done! believe it or not. because we took the log as the first step

  75. anonymous
    • 5 years ago
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    i cud not pick that up!!!!! that ln(sin(x) ) was by a rule??????????

  76. anonymous
    • 5 years ago
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    we did not find the derivative of the original thing, we found the derivative of the log of the original thing.

  77. anonymous
    • 5 years ago
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    so wat next?

  78. anonymous
    • 5 years ago
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    ok i will say quickly what we do next because it is easy. then i will give an explanation. next we simply get our answer by multiplying what we got above by the original function. done.

  79. anonymous
    • 5 years ago
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    CONFUSION HERE IN WAT U JUST SAID (short circuited)!!!!!!!!!!!!!!!!!!!!!!!!

  80. anonymous
    • 5 years ago
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    answer is... \[(sin(x))^{x^2}\times (2xln(sin(x))+\frac{x^2cos(x)}{sin(x)})\]

  81. anonymous
    • 5 years ago
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    gotta go!!!!!!!

  82. anonymous
    • 5 years ago
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    the first part is the original function you were to differentiate. the second part was the derivative of the log of that good luck

  83. anonymous
    • 5 years ago
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    i can 't understand the final step on how to obtain the answer

  84. anonymous
    • 5 years ago
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    ok final step. don't forget that the first step was taking the log. so we took the derivative of \[ln(f(x))\] not \[f(x)\] by the chain rule the derivative of \[ln(f(x))\] is \[\frac{1}{f(x)}\times f'(x)=\frac{f'(x)}{f(x)}\] that is \[\frac{d}{dx}ln(f(x))=\frac{f'(x)}{f(x)}\] solving this equation for \[f'(x)\] gives \[f'(x)=f(x)\times \frac{d}{dx}ln(f(x))\] in english you can find the derivative by first taking the log of your function, then differentiating the log of your function, and then multiplying by the original function. hope this helps

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