Find the derivative using logarithmic differentiation method $\sqrt{\frac{x^3-y^3}{x+y}\frac{x^2+2xy+y^2}{x^2+xy+y^2}}\frac{x^2-y^2}{4}$

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Final answer to the problem

$\frac{y^{\prime}}{y}=\frac{1}{x^2-y^2}\left(2x-2y\cdot y^{\prime}\right)+\frac{1}{2}\left(\frac{1}{x^3-y^3}\left(3x^{2}-3y^{2}y^{\prime}\right)+\frac{1}{x^2+2xy+y^2}\left(2x+2\left(y+xy^{\prime}\right)+2y\cdot y^{\prime}\right)+\frac{-1}{x+y}\left(1+y^{\prime}\right)+\frac{-1}{x^2+xy+y^2}\left(2x+y+xy^{\prime}+2y\cdot y^{\prime}\right)\right)$
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Step-by-step Solution

How should I solve this problem?

  • Find the derivative using logarithmic differentiation
  • Find the derivative using the definition
  • Find the derivative using the product rule
  • Find the derivative using the quotient rule
  • Find the derivative
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Integrate by substitution
  • Integrate by parts
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Simplify the derivative by applying the properties of logarithms

$\frac{d}{dx}\left(\frac{\left(x^2-y^2\right)\sqrt{\frac{\left(x^3-y^3\right)\left(x^2+2xy+y^2\right)}{\left(x+y\right)\left(x^2+xy+y^2\right)}}}{4}\right)$

Learn how to solve differential calculus problems step by step online.

$\frac{d}{dx}\left(\frac{\left(x^2-y^2\right)\sqrt{\frac{\left(x^3-y^3\right)\left(x^2+2xy+y^2\right)}{\left(x+y\right)\left(x^2+xy+y^2\right)}}}{4}\right)$

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Learn how to solve differential calculus problems step by step online. Find the derivative using logarithmic differentiation method ((x^3-y^3)/(x+y)(x^2+2xyy^2)/(x^2+xyy^2))^(1/2)(x^2-y^2)/4. Simplify the derivative by applying the properties of logarithms. To derive the function \frac{\left(x^2-y^2\right)\sqrt{\frac{\left(x^3-y^3\right)\left(x^2+2xy+y^2\right)}{\left(x+y\right)\left(x^2+xy+y^2\right)}}}{4}, use the method of logarithmic differentiation. First, assign the function to y, then take the natural logarithm of both sides of the equation. Apply natural logarithm to both sides of the equality. Apply logarithm properties to both sides of the equality.

Final answer to the problem

$\frac{y^{\prime}}{y}=\frac{1}{x^2-y^2}\left(2x-2y\cdot y^{\prime}\right)+\frac{1}{2}\left(\frac{1}{x^3-y^3}\left(3x^{2}-3y^{2}y^{\prime}\right)+\frac{1}{x^2+2xy+y^2}\left(2x+2\left(y+xy^{\prime}\right)+2y\cdot y^{\prime}\right)+\frac{-1}{x+y}\left(1+y^{\prime}\right)+\frac{-1}{x^2+xy+y^2}\left(2x+y+xy^{\prime}+2y\cdot y^{\prime}\right)\right)$

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Function Plot

Plotting: $\frac{y^{\prime}}{y}=\frac{1}{x^2-y^2}\left(2x-2y\cdot y^{\prime}\right)+\frac{1}{2}\left(\frac{1}{x^3-y^3}\left(3x^{2}-3y^{2}y^{\prime}\right)+\frac{1}{x^2+2xy+y^2}\left(2x+2\left(y+xy^{\prime}\right)+2y\cdot y^{\prime}\right)+\frac{-1}{x+y}\left(1+y^{\prime}\right)+\frac{-1}{x^2+xy+y^2}\left(2x+y+xy^{\prime}+2y\cdot y^{\prime}\right)\right)$

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a
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g
m
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u
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x
y
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.
(◻)
+
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×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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Main Topic: Differential Calculus

The derivative of a function of a real variable measures the sensitivity to change of a quantity (a function value or dependent variable) which is determined by another quantity (the independent variable). Derivatives are a fundamental tool of calculus.

Used Formulas

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