Derivative of the square root of xy

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SUMMARY

The derivative of the square root of the product of two variables, \( \sqrt{xy} \), is calculated using the chain rule and implicit differentiation. The correct derivative is expressed as \( \frac{1}{2}(xy)^{-1/2}(y + x\frac{dy}{dx}) \). To simplify, it can be rewritten as \( \frac{1}{2}x^{1/2}y^{3/2} + x^{3/2}y^{1/2}\frac{dy}{dx} \). The discussion emphasizes the importance of applying the product rule and correctly handling half powers during differentiation.

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  • Understanding of calculus concepts, specifically differentiation
  • Familiarity with the chain rule and product rule in calculus
  • Knowledge of implicit differentiation techniques
  • Basic algebraic manipulation of exponents and roots
NEXT STEPS
  • Study the application of the chain rule in more complex functions
  • Learn about implicit differentiation in detail
  • Explore the product rule and its applications in calculus
  • Practice problems involving derivatives of products and roots
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Students studying calculus, particularly those focusing on differentiation techniques, and educators looking for examples of implicit differentiation and product rule applications.

brambleberry
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Homework Statement



What is the deriv. of the square root of (xy)?

Homework Equations





The Attempt at a Solution



I used the chain rule:

(1/2)(xy)^(-1/2) times (y + x(dy/dx))

i am unsure on how to distribute this correctly
 
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The derivative of (xy)^{1/2} with respect to x, and y is a function of x?

If that is the question, then yes that is correct. I don't know what you mean "how to distribute this correctly". The distributive law is the distributive law: a(b+ c)= ab+ ac.
Is it the half powers that concern you? (xy)^{1/2}x= (x^{1/2})(x)(y^{1/2}= x^{3/2}y^{1/2} and (xy)^{1/2}y= (x^{1/2})(y^{1/2})y= x^{1/2}y^{3/2}.

(1/2)(xy)^{1/2}[y+ x dy/dx]= (1/2)x^{1/2}y^{3/2}+ x^{3/2}y^{1/2} dy/dx
 
Are you trying to do implicit differentiation? If so treat

\sqrt{xy} = \sqrt{x}\sqrt{y}

Then use the product rule, just remember when you differentiate \sqrt{y} to multiply by y'.
 

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