Partial Derivative of x^y: How to Find the First Partial Derivatives?

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SUMMARY

The discussion focuses on finding the first partial derivatives of the functions f(x,y) = x^y and u = x^(y/z). The correct first partial derivatives are established as f_x = y*x^(y-1) and f_y = ln(x)*x^y. For the second function, u_x is confirmed as (y/z)*x^((y/z)-1), while u_y requires the application of the chain rule, leading to u_y = (ln(x)/z)*x^(y/z) and u_z = (y*ln(x)/z)*x^(y/z). The importance of treating x as a constant and using the exponential function base is emphasized for accurate differentiation.

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Homework Statement



Find the first partial derivatives of:

1. f(x,y) = x^y
2. u = x^(y/z)


Homework Equations





The Attempt at a Solution



f_x = y*x^(y-1)
f_y = lnx?

u_x = (y/z)*x^((y/z)-1)
u_y = lnx/z?
u_z = ylnx/z?

I'm not really sure how to do these right. =/ I would really appreciate any help.
 
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Your f_x is right. Your f_y is not. Look at x as a constant in this one and look up the derivative of an exponential of arbitrary base formula.

Your u_x is right.
Your u_y again should be treated as an exponential function base x.
Your u_z should as well with an additional application of the chain rule.
 
Thank you!
 
Don't forget to use the chain-rule.

For the y derivative of x^y:

Let x = k, a constant.

f(y) = k^y

Natural log of both sides gives:

ln(f(y)) = ln(k^y)

ln(f(y)) = yln(k)

Differentiating...

f'(y)/f(y) = ln(k)

f'(y) = f(y)ln(k)

Since f(y) = k^y, you now have:

f'(y) = ln(k)k^y

Substituting for x...

f_y = ln(x)x^y
 

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