Derivative of the square root of xy

In summary, the conversation is about finding the derivative of the square root of xy using the chain rule and implicit differentiation. The final answer is (1/2)x^(1/2)y^(3/2) + x^(3/2)y^(1/2)dy/dx.
  • #1
brambleberry
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0

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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  • #2
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]
 
  • #3
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'.
 

FAQ: Derivative of the square root of xy

What is the derivative of the square root of xy?

The derivative of the square root of xy is equal to (1/2)(x^(-1/2)y + x^(1/2)y^(-1)).

How do you find the derivative of the square root of xy?

To find the derivative of the square root of xy, use the power rule and the chain rule. First, rewrite the expression as (xy)^1/2. Then, apply the power rule to get (1/2)(xy)^(-1/2). Finally, use the chain rule to multiply by the derivative of the inside function, which is (y + x). The final derivative is (1/2)(x^(-1/2)y + x^(1/2)y^(-1)).

Can the derivative of the square root of xy be simplified?

Yes, the derivative of the square root of xy can be simplified using the power rule and the chain rule. The simplified form is (1/2)(x^(-1/2)y + x^(1/2)y^(-1)).

What is the significance of the derivative of the square root of xy?

The derivative of the square root of xy represents the rate of change of this function with respect to its variables x and y. It can be used to find the slope of a tangent line at a specific point on the curve, as well as to determine maximum and minimum values of the function.

Are there any real-life applications of the derivative of the square root of xy?

Yes, the derivative of the square root of xy has many real-life applications, particularly in physics and engineering. For example, it can be used to calculate the velocity of an object in projectile motion, or to determine the optimal dimensions of a structure given certain constraints.

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