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MTALHAHASSAN2

  • one year ago

Use the product rule and the power of a function rule to differenite the following functions. Do not simplify. b) (3x^2+4)(3+x^3)^5

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  1. MTALHAHASSAN2
    • one year ago
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    @Shalante

  2. anonymous
    • one year ago
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    This can be solved using the product rule Its just 2 function multiplying together: f(x) and g(x) Lets call \[f(x)=3x^2+4\] \[g(x)=(3+x^3)^5\] Try to find the derivative of both by yourself f'(x)=? g'(x)=? then use this product rule formula f'(x)g(x)+f(x)g'(x)

  3. MTALHAHASSAN2
    • one year ago
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    y=(3x^2+4)(3+x^3)^5 =2(3) (3+x^3)^5 + (3x^2+4) 15(3+x^2)

  4. MTALHAHASSAN2
    • one year ago
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    is it right???

  5. Jhannybean
    • one year ago
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    \[f(x) = (3x^2+4)(3+x^3)^5\]\[f(x) = 3x^2+4 ~,~ g(x) = (3+x^2)^5\]

  6. anonymous
    • one year ago
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    Derivative of 3x^2 is not 6 Derivative (3+x^3)^5 is not 15(3+x^2)

  7. Jhannybean
    • one year ago
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    \[\frac{d}{dx} (f(x)) = \frac{d}{dx}(3x^2+4) \cdot (3+x^3)^5 +\frac{d}{dx}((3+x^3)^5) \cdot (3x^2+4)\]

  8. anonymous
    • one year ago
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    You forgot the x in the first one You forgot the x^2 in the second one.

  9. Jhannybean
    • one year ago
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    Remember to apply the chainrule.

  10. Jhannybean
    • one year ago
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    \[\frac{d}{dx}(f(g(x)) = f'(g(x)) \cdot g'(x)\]

  11. anonymous
    • one year ago
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    He did apply the chain rule. He forgot the x on both cases but not the numbers.

  12. Jhannybean
    • one year ago
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    Ahh ok

  13. MTALHAHASSAN2
    • one year ago
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    =2(3x)(3+x^3)^5+(3x^2+4)5(3x^2)(3+x^3)^4 =6x(3+x^3)^5+15x^2(3x^2+4)(3+x^3)^4

  14. anonymous
    • one year ago
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    Yes.

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