Practice Differentiate 3

harpazo

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Differentiate.

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TheJason

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\(\displaystyle y = \dfrac{2x - \cos\,x }{x^{2}}\)

\(\displaystyle y'= \dfrac{(x^{2})(2 - (-\sin\,x)) - (2x)(2x - \cos\,x) }{(x^{2})^{2}}\)

\(\displaystyle y'= \dfrac{(x^{2})(2 + \sin\,x) - (2x)(2x - \cos\,x) }{(x^{2})^{2}}\)

\(\displaystyle y' = \dfrac{2x^{2} + \sin^{3}x - 4x^{2} - 2\cos^{2}x}{x^{4}}\)

\(\displaystyle y' = \dfrac{-2x^{2} + \sin^{3}x - 2\cos^{2}x}{x^{4}}\)
 
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MarkFL

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Jason, you've made an error in your math:

\(\displaystyle x^2\cdot\sin(x)\ne\sin^3(x)\)

\(\displaystyle x\cdot\cos(x)\ne\cos^2(x)\)
 
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TheJason

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Jason, you've made an error in your math:

\(\displaystyle x^2\cdot\sin(x)\ne\sin^3(x)\)

\(\displaystyle x\cdot\cos(x)\ne\cos^2(x)\)
\(\displaystyle y = \dfrac{2x - \cos\,x }{x^{2}}\)

\(\displaystyle y'= \dfrac{(x^{2})(2 - (-\sin\,x)) - (2x)(2x - \cos\,x) }{(x^{2})^{2}}\)

\(\displaystyle y'= \dfrac{(x^{2})(2 + \sin\,x) - (2x)(2x - \cos\,x) }{(x^{2})^{2}}\)

\(\displaystyle y'= \dfrac{2x^{2} + x^{2}\sin\,x - 4x^{2} + 2x\cos\,x }{x^{4}}\)

\(\displaystyle y'= \dfrac{-2x^{2} + x^{2}\sin\,x + 2x\cos\,x }{x^{4}}\)
 

MarkFL

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Yes, that appear to be correct...I would go on to divide by numerator and denominator by \(x\). :)
 
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harpazo

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Simplify as much as possible.