- #1

- 52

- 0

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.

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- Thread starter physicsed
- Start date

- #1

- 52

- 0

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.

- #2

- 52

- 0

or maybe the answer is

[tex] y=-2cosx [/tex]

[tex] y=-2cosx [/tex]

- #3

- 334

- 0

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]

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

- 690

- 6

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.

[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

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