MTALHAHASSAN2
  • MTALHAHASSAN2
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
Mathematics
schrodinger
  • schrodinger
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MTALHAHASSAN2
  • MTALHAHASSAN2
anonymous
  • anonymous
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)
MTALHAHASSAN2
  • MTALHAHASSAN2
y=(3x^2+4)(3+x^3)^5 =2(3) (3+x^3)^5 + (3x^2+4) 15(3+x^2)

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MTALHAHASSAN2
  • MTALHAHASSAN2
is it right???
Jhannybean
  • Jhannybean
\[f(x) = (3x^2+4)(3+x^3)^5\]\[f(x) = 3x^2+4 ~,~ g(x) = (3+x^2)^5\]
anonymous
  • anonymous
Derivative of 3x^2 is not 6 Derivative (3+x^3)^5 is not 15(3+x^2)
Jhannybean
  • Jhannybean
\[\frac{d}{dx} (f(x)) = \frac{d}{dx}(3x^2+4) \cdot (3+x^3)^5 +\frac{d}{dx}((3+x^3)^5) \cdot (3x^2+4)\]
anonymous
  • anonymous
You forgot the x in the first one You forgot the x^2 in the second one.
Jhannybean
  • Jhannybean
Remember to apply the chainrule.
Jhannybean
  • Jhannybean
\[\frac{d}{dx}(f(g(x)) = f'(g(x)) \cdot g'(x)\]
anonymous
  • anonymous
He did apply the chain rule. He forgot the x on both cases but not the numbers.
Jhannybean
  • Jhannybean
Ahh ok
MTALHAHASSAN2
  • MTALHAHASSAN2
=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
anonymous
  • anonymous
Yes.

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