Solving a derivative: y = x e^ x csc x
- Thread starter grapz
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Not quite. When you're using the product rule, you have to split y into two expressions and use the rule accordingly (since the product rule deals with two functions). So say that this equation was expressed as y = f(x)g(x). You could let f(x) = xe<sup>x</sup> and g(x) = cscx or f(x) = x and g(x) = e<sup>x</sup>cscx. Either way, the product rule will work the same and give you equivalent answers for the derivative.
y' = f'(x)g(x) + f(x)g'(x)
y' = (xe<sup>x</sup>)'(cscx) + (xe<sup>x</sup>)(cscx)'
As you can see, f'(x) requires the additional product rule which you could probably manage yourself.
y' = f'(x)g(x) + f(x)g'(x)
y' = (xe<sup>x</sup>)'(cscx) + (xe<sup>x</sup>)(cscx)'
As you can see, f'(x) requires the additional product rule which you could probably manage yourself.
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Re: Solving a derivative
y' =( x e ^ x ) ( csc x )
That is just wrong. That it 'y'. Why do you have it listed as "y-prime"? I know you don't mean that, but that is what you wrote.
y' = ( e ^ x ( x) + e ^ x ) ( csc x )
Again with the y-prime as in intermediate result. No good.
Be more careful.
y = x e^ x csc x
Take f(x) = x e^x
Then f'(x) = e ^ x ( x) + e ^ x
You had that, but with bad notation.
y = f(x) csc(x)
By the product rule
y' = f(x)*( - cot x csc x ) + f'(x)*csc(x)
This is were you went really off. Rather than insert the function, f(x) and its derivative, f'(x), you managed to use f'(x) and f"(x). You should have gotten:
y' = (x e^x)*( - cot x csc x ) + (e ^ x ( x) + e ^ x)*csc(x)
This is much simpler that your expression. Also, this can be simplified substantially.
y' = (e^x)(csc(x))(x + 1 - x cot(x))
You have the right idea, but you have confused yourself. Mostly, you let the notation get out of hand.grapz said:
y' =( x e ^ x ) ( csc x )
That is just wrong. That it 'y'. Why do you have it listed as "y-prime"? I know you don't mean that, but that is what you wrote.
y' = ( e ^ x ( x) + e ^ x ) ( csc x )
Again with the y-prime as in intermediate result. No good.
Be more careful.
y = x e^ x csc x
Take f(x) = x e^x
Then f'(x) = e ^ x ( x) + e ^ x
You had that, but with bad notation.
y = f(x) csc(x)
By the product rule
y' = f(x)*( - cot x csc x ) + f'(x)*csc(x)
This is were you went really off. Rather than insert the function, f(x) and its derivative, f'(x), you managed to use f'(x) and f"(x). You should have gotten:
y' = (x e^x)*( - cot x csc x ) + (e ^ x ( x) + e ^ x)*csc(x)
This is much simpler that your expression. Also, this can be simplified substantially.
y' = (e^x)(csc(x))(x + 1 - x cot(x))