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Differentiation of a quotient of two function?
Question: y=Cosx/x^4
Answer : dy/dx=x^4(-sinx)-cosx(4x^3) / x^8
= -xsinx - 4cosx / x^5
Okay I have a question regarding the final part of the answer -xsinx - 4cosx / x^5 , how is this worked out. I got to the point of dy/dx=x^4(-sinx)-cosx(4x^3) / x^8 but after that I'm confused. Why does x^4(-sinx) become -xsinx on the last step, similarly -cosx(4x^3) to - 4cosx and x^8 to x^5. confused and I need some help. Thank you.
okay I think I just figured it out...after first step I did this : x^3(-xcosx-4cosx) / x^8, note: this is same as x^4(-sinx)-cosx(4x^3) / x^8 , so I subtracted x^3 by the denominator x^8 which gave me x^5...the x^3 from the top cancels out...Correct me if I'm wrong, please. thank you.
2 Answers
- Anonymous9 years agoFavorite Answer
y = cos(x) / x^4
Can be written as:
y = x^(-4) * cos(x)
Using the product rule:
dy/dx = d/dx (x^(-4)) * cos(x) + x^(-4) * d/dx (cos(x))
dy/dx = -4x^(-5) * cos(x) + x^(-4) * -sin(x)
dy/dx = -4cos(x) / (x^5) - sin(x) / (x^4)
Make them the same denominator by multiplying the second fraction by x/x:
dy/dx = (-4cos(x) - xsin(x)) / (x^5)
I hope this clears things up :)
- 9 years ago
Ok so when you have dy/dx=x^4(-sinx)-cosx(4x^3) / x^8, multiply each term in the equation by x^-3 (ie drop the power of each x by 3)
dy/dx=x^1(-sinx)-cosx(4x^0) / x^5
= -xsinx - 4cosx / x^5
