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# Find the implicit derivative $\frac{d}{dx}\left(x^xy=\sqrt[3]{\frac{x\left(x+1\right)\left(x-2\right)}{\left(x^2+1\right)\left(2x+3\right)}}\right)$

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##  Final answer to the problem

$\left(\ln\left(x\right)+1\right)x^xy+x^xy^{\prime}=\frac{\left(\left(x+1\right)\left(x-2\right)+x\left(x-2+x+1\right)\right)\left(x^2+1\right)\left(2x+3\right)+\left(-x-1\right)x\left(x-2\right)\left(2x\left(2x+3\right)+2\left(x^2+1\right)\right)}{3\left(x^2+1\right)^2\left(2x+3\right)^2}\sqrt[3]{\left(\frac{\left(x^2+1\right)\left(2x+3\right)}{x\left(x+1\right)\left(x-2\right)}\right)^{2}}$
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##  Step-by-step Solution 

How should I solve this problem?

• Choose an option
• Find the derivative using the definition
• Find the derivative using the product rule
• Find the derivative using the quotient rule
• Find the derivative using logarithmic differentiation
• Find the derivative
• Integrate by partial fractions
• Product of Binomials with Common Term
• FOIL Method
• Integrate by substitution
Can't find a method? Tell us so we can add it.
1

Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

$\frac{d}{dx}\left(x^xy\right)=\frac{d}{dx}\left(\sqrt[3]{\frac{x\left(x+1\right)\left(x-2\right)}{\left(x^2+1\right)\left(2x+3\right)}}\right)$

Learn how to solve implicit differentiation problems step by step online.

$\frac{d}{dx}\left(x^xy\right)=\frac{d}{dx}\left(\sqrt[3]{\frac{x\left(x+1\right)\left(x-2\right)}{\left(x^2+1\right)\left(2x+3\right)}}\right)$

Learn how to solve implicit differentiation problems step by step online. Find the implicit derivative d/dx(x^xy=((x(x+1)(x-2))/((x^2+1)(2x+3)))^(1/3)). Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=. The derivative of the linear function is equal to 1. The power rule for differentiation states that if n is a real number and f(x) = x^n, then f'(x) = nx^{n-1}.

##  Final answer to the problem

$\left(\ln\left(x\right)+1\right)x^xy+x^xy^{\prime}=\frac{\left(\left(x+1\right)\left(x-2\right)+x\left(x-2+x+1\right)\right)\left(x^2+1\right)\left(2x+3\right)+\left(-x-1\right)x\left(x-2\right)\left(2x\left(2x+3\right)+2\left(x^2+1\right)\right)}{3\left(x^2+1\right)^2\left(2x+3\right)^2}\sqrt[3]{\left(\frac{\left(x^2+1\right)\left(2x+3\right)}{x\left(x+1\right)\left(x-2\right)}\right)^{2}}$

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

SnapXam A2

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x
y
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(◻)
+
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×
◻/◻
/
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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

###  Main Topic: Implicit Differentiation

Implicit differentiation makes use of the chain rule to differentiate implicitly defined functions. For differentiating an implicit function y(x), defined by an equation R(x, y) = 0, it is not generally possible to solve it explicitly for y(x) and then differentiate. Instead, one can differentiate R(x, y) with respect to x and y and then solve a linear equation in dy/dx for getting explicitly the derivative in terms of x and y.