Practice Differentiate 3

harpazo

Pure Mathematics
Mar 20, 2018
5,788
361
83
NYC
Differentiate.

sDraw_2018-10-09_19-50-35.png
 

Jason

Administrator
Staff member
Administrator
Moderator
Jan 25, 2018
2,520
547
113
jasonyankee.info
\(\displaystyle y = \dfrac{2x - \cos\,x }{x^{2}}\)

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

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

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

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

MarkFL

La Villa Strangiato
Staff member
Administrator
Moderator
Math Helper
Jan 25, 2018
3,522
4,289
113
St. Augustine
Jason, you've made an error in your math:

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

\(\displaystyle x\cdot\cos(x)\ne\cos^2(x)\)
 
  • Like
Reactions: anemone

Jason

Administrator
Staff member
Administrator
Moderator
Jan 25, 2018
2,520
547
113
jasonyankee.info
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

La Villa Strangiato
Staff member
Administrator
Moderator
Math Helper
Jan 25, 2018
3,522
4,289
113
St. Augustine
Yes, that appear to be correct...I would go on to divide by numerator and denominator by \(x\). :)
 
  • Like
Reactions: anemone and harpazo

harpazo

Pure Mathematics
Mar 20, 2018
5,788
361
83
NYC
Simplify as much as possible.