Derivative of the square root of xy

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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 [itex](xy)^{1/2}[/itex] 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? [itex](xy)^{1/2}x= (x^{1/2})(x)(y^{1/2}= x^{3/2}y^{1/2}[/itex] and [itex](xy)^{1/2}y= (x^{1/2})(y^{1/2})y= x^{1/2}y^{3/2}[/itex].

[itex](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[/itex]
 
Are you trying to do implicit differentiation? If so treat

[tex]\sqrt{xy} = \sqrt{x}\sqrt{y}[/tex]

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