amy0799
  • amy0799
The function f and g are differentiable at x=10 and x=20 and f(g(x))=x^2. if f(10)=5, f'(10)=4, f'(20)=-5, and g(10)=20, what is the value of g'(10)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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freckles
  • freckles
have you used chain rule to differentiate f(g(x)) yet?
amy0799
  • amy0799
I learned how to do the chain rule
freckles
  • freckles
ok then what is \[\frac{d}{dx}f(g(x))=?\]

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amy0799
  • amy0799
f(x)g'(x)+g(x)f'(x)
freckles
  • freckles
that is the product rule we don't have f*g we have f composed with g
freckles
  • freckles
this is why I asked you to use chain rule to differentiate f(g(x))
amy0799
  • amy0799
I thought u use the chain rule when its something to a power
freckles
  • freckles
power rule is what you use when you have a constant power
freckles
  • freckles
chain rule is what you use when you have a function inside a function
amy0799
  • amy0799
ooh ok. So how would I do the chain rule for this? would it be f'(g(x))*g'(x)?
freckles
  • freckles
that is right
freckles
  • freckles
\[f(g(x))=x^2 \\ \text{ differentiating both sides } \\ f'(g(x)) \cdot g'(x)=2x\]
freckles
  • freckles
not enter in 10 for x
freckles
  • freckles
\[f'(g(10)) \cdot g'(10)=2(10)\]
freckles
  • freckles
you are given g(10)
freckles
  • freckles
g(10)=20 right?
freckles
  • freckles
so replace g(10) with 20 \[f'(g(10)) \cdot g'(10)=2(10)\] \[f'(20) \cdot g'(10)=2(10)\]
freckles
  • freckles
see if you can finish the rest
amy0799
  • amy0799
g'(10)=25?
freckles
  • freckles
\[f'(20)=-5 \\ \text{ so we have } \\ -5 \cdot g'(10)=2(10)\]
freckles
  • freckles
hmmm how did you get g'(10)=25?
amy0799
  • amy0799
oh I added, oops
freckles
  • freckles
you do know the operation between -5 and g'(10) is multiplication and not addition :p
amy0799
  • amy0799
g'(10)=-4?
freckles
  • freckles
20/-5 is -4 good worrk
amy0799
  • amy0799
thank you!
amy0799
  • amy0799
I have another question, do u mind helping me till?
freckles
  • freckles
I can take a look
amy0799
  • amy0799
In the table below, the values of f(x), g(x), f'(x) and g'(x) are given 2 values of x. if y =[f(2x)+g(x)]^2, find y'(3)
freckles
  • freckles
well we know we are going to have differentiate since we want to find y'(of something)
amy0799
  • amy0799
|dw:1441749521965:dw|
freckles
  • freckles
again here you will have to use chain rule this chain rule does come with power rule because we have have a constant power we will also have to use sum rule then chain rule again
amy0799
  • amy0799
2(f(2x)+g(x))*(f'(2x)+g'(x))
freckles
  • freckles
ok I only have one complaint...
freckles
  • freckles
let's look at f(2x)
freckles
  • freckles
notice the inside function is 2x
freckles
  • freckles
the outside function is f( )
freckles
  • freckles
when you find the derivative of f(2x) w.r.t. x you should get 2*f'(2x) derivative of inside time derivative of outside
freckles
  • freckles
\[y(x)=[f(2x)+g(x)]^2 \\ y'(x)=2[f(2x)+g(x)] \cdot [2 f'(2x)+g'(x)]\]
freckles
  • freckles
now we want to find y'(3)
freckles
  • freckles
so replace x with 3
freckles
  • freckles
\[\\ y'(3)=2[f(2\cdot 3)+g(3)] \cdot [2 f'(2 \cdot 3)+g'(3)]\]
amy0799
  • amy0799
y'(3)=112
freckles
  • freckles
\[f(6)=? \\ g(3)=? \\ f'(6)=? \\ g'(3)=?\] you found these values from the table ?
freckles
  • freckles
\[f(6)=-3 \\ g(3)=-1 \\ f'(6)=-2 \\ g'(3)=-5\] this is what I see for those values is that what you also see
amy0799
  • amy0799
oh I know what I did wrong. Hold on
amy0799
  • amy0799
y'(3)=72?
freckles
  • freckles
purrrfect (you know like a cat)
amy0799
  • amy0799
haha thank you so much for the help!
freckles
  • freckles
np I hope everything makes more sense
amy0799
  • amy0799
Yup it does thanks to your help :D
freckles
  • freckles
np I must go now good luck with calculus
amy0799
  • amy0799
ok. thanks, I need all the luck I can get haha

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