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JayCT Group Title

find the derivative of 2x -5sqrt(x) using rules of differentiations please

  • one year ago
  • one year ago

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  1. shashi20008 Group Title
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    2 - 10x

    • one year ago
  2. JayCT Group Title
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    can you show steps pls

    • one year ago
  3. shashi20008 Group Title
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    yeah sure, \[\frac{ d}{ dx }(2x-5x ^{2})\] \[=\frac{ d}{ dx }(2x)-\frac{ d}{ dx }(5x ^{2})\] \[=2 - 5\frac{ d }{ dx }(x ^{2})\] \[=2 - 5 * 2x\] \[= 2- 5x\]

    • one year ago
  4. JayCT Group Title
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    ok thx very much. i have another one similar f(t)= 2t^2 - sqrt(t^3)

    • one year ago
  5. shashi20008 Group Title
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    its quite similar, you have to do same thing. simply remember, while differentiating some variable exponent some constant number, you just have to multiply the exponent to coefficient of the term and then write the variable after decreasing the exponent by one. Do this for each term in an expression and you would be done :) as for your ques, \[f'(x) = \frac{ d }{ dx }(2t ^{2} - \sqrt{t ^{3}})\] \[= \frac{ d }{ dx }(2t ^{2}) - \frac{ d }{ dx }(\sqrt{t ^{3}})\] \[=2 \frac{ d }{ dx }(t ^{2}) - \frac{ d }{ dx }(t ^{\frac{ 3 }{ 2 }})\] \[=2 *2t - \frac{ 3 }{ 2 }(t ^{\frac{ 1 }{ 2 }})\] \[=4t - \frac{ 3 }{ 2 }(\sqrt{t})\]

    • one year ago
  6. JayCT Group Title
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    wow thanks very much. i have few other questions, dont know if you would mind helping me?

    • one year ago
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