Easy chain rule for you guys

  • Thread starter physicsed
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  • #1
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[tex] y=2x^{sinx}[/tex]

i know i should use the product rule within a chain rule. but how can i use chain rule with sinx

is the anwser

[tex] y=-2x^{cosx}[/tex]

can anyone give me pointer to this easy problem and tell if am forgetting something.
 

Answers and Replies

  • #2
52
0
or maybe the answer is

[tex] y=-2cosx [/tex]
 
  • #3
334
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That definitely is not the answer. You need to realize what you're doing when you apply the chain rule. First of all, you are attemping to find [itex]\frac{dy}{dx}[/itex], which you have not indicated. Show your work and maybe we can help, but you first have to transform the equation into something you can work with (i.e. get rid of the exponent). How would you accomplish this?

Edit: Is your equation [itex]y=2xsin(x)[/itex] or [itex]y=2x^{sinx}[/itex]?
 
  • #4
52
0
[itex]\frac{dy}{dx}[/itex] [tex] {y= 2x^{sinx}} [/tex]

product rule
[tex] f= 2 f'= 0
g= x g'=? [/tex]
chain rule
[tex] f= x f'= 1
g=? g'=? [/tex]
is the anser
[tex]
y'=-2x^{cosx} [/tex]
 
Last edited:
  • #5
You're making the assumption that:

[tex]\frac{d}{dx}\;2x^{sin(x)}=-2x^{cos(x)}\rightarrow (f \circ g)' = f'\circ g\cdot g'=-2x^{cos(x)}[/tex]

It doesn't.

You're using the chain rule but incorrectly, use the product rule on the results.
 
Last edited:
  • #6
cristo
Staff Emeritus
Science Advisor
8,107
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To evaluate this derivative I would take logarithms of both sides to get ln(y)=sinx.ln(2x) and then differentiate implicitly.
 

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