anonymous
  • anonymous
Let y =(root5(x^8+5)). Find values A and B so that (dy)/(dx) = A(x^8+5)^B(C x^D)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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myininaya
  • myininaya
\[y=\sqrt[5]{x^8+5}\] \[y=(x^8+5)^\frac{1}{5}\] Use the chain rule to find y'
anonymous
  • anonymous
what if I need to find A,B,C, and D? what do I plug in?
myininaya
  • myininaya
You need to find y' first to find A,B,C,and D

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More answers

anonymous
  • anonymous
I havent learned the chain rule yet
myininaya
  • myininaya
\[[f(g(x))]'=f'(g(x)) \cdot g'(x) \text{ chain rule }\]
myininaya
  • myininaya
Well you need the chain rule to do this problem.
myininaya
  • myininaya
what is g(x) and what is f(x) if \[f(g(x))=(x^8+5)^\frac{1}{5}\]
anonymous
  • anonymous
g(x)= 8x f(x)= 5^1/5
myininaya
  • myininaya
No g(x) is the inside function and f(x) is the outside function
anonymous
  • anonymous
oh ok g(x)= 8x^7
myininaya
  • myininaya
\[g(x)=x^8+5 ; f(x)=x^\frac{1}{5}\] \[g'(x)=8x^7+0 ; f(x)=\frac{1}{5}x^{\frac{1}{5}-1}=\frac{1}{5}x^\frac{-4}{5}\]
anonymous
  • anonymous
so I plug this in in the equation now
myininaya
  • myininaya
this one \[y'=[f(g(x))]'=f'(g(x)) \cdot g'(x) \text{ chain rule } \] yes
anonymous
  • anonymous
1/5x^-4/5
myininaya
  • myininaya
\[y'=f'(x^8+5) \cdot 8x^7\]
anonymous
  • anonymous
now just multiply, right?
myininaya
  • myininaya
you need to plug x^8+5 into f'
anonymous
  • anonymous
oh ok
myininaya
  • myininaya
\[g'(x)=8x^7+0 ; f'(x)=\frac{1}{5}x^{\frac{1}{5}-1}=\frac{1}{5}x^\frac{-4}{5}\]
myininaya
  • myininaya
f' is that second one I wrote down just now
myininaya
  • myininaya
\[y'=f'(x^8+5) \cdot 8x^7=\frac{1}{5}(x^8+5)^\frac{-4}{5} \cdot 8x^7\]
myininaya
  • myininaya
now you should be able to put it in the form above to get your A,B,C, and D
anonymous
  • anonymous
i'll try it
myininaya
  • myininaya
actually it is already in that form
anonymous
  • anonymous
so just look for the A,B,C,D then right?
myininaya
  • myininaya
yes

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