Thread: Partial Derivatives

Given the following function, find f_(xx), f_(xy), f_(yyx) [Note: f_(xx), for instance, is f subscript xx].

f(x,y) = e^(4x) - sin(y^2) - sqrt(xy)

Answers in book:

f_(xx) = 16e^(4x) + (1/4)*sqrt(y/(x^3))
f_(xy) = -1/(4*sqrt(xy))
f_(yyx) = 1/(8*sqrt(x^3*y^3))

Originally Posted by Ideasman

Given the following function, find f_(xx), f_(xy), f_(yyx) [Note: f_(xx), for instance, is f subscript xx].

f(x,y) = e^(4x) - sin(y^2) - sqrt(xy)

Answers in book:

f_(xx) = 16e^(4x) + (1/4)*sqrt(y/(x^3))
f_(xy) = -1/(4*sqrt(xy))
f_(yyx) = 1/(8*sqrt(x^3*y^3))

I'll do f_(xx), you can do the rest.

To get a partial derivative with respect to x, treat all other variables as constants. So given:
f = e^(4x) - sin(y^2) - sqrt(xy)

f_(x) = 4e^(4x) - (1/2)*(1/sqrt(xy))*y

f_(x) = 4e^(4x) - (1/2)sqrt(y/x)

To proceed it is simpler to look at this as:
f_(x) = 4e^(4x) - (1/2)sqrt(y)*x^(-1/2)

So:
f_(xx) = 16e^(4x) - (1/2)sqrt(y)*(-1/2)x^(-3/2)

Which is the same as:
f_(xx) = 16e^(4x) + (1/4)sqrt(y/x^3)

The other two are done the same way, though they require a bit more work.

-Dan

Originally Posted by topsquark

I'll do f_(xx), you can do the rest.

To get a partial derivative with respect to x, treat all other variables as constants. So given:
f = e^(4x) - sin(y^2) - sqrt(xy)

f_(x) = 4e^(4x) - (1/2)*(1/sqrt(xy))*y

f_(x) = 4e^(4x) - (1/2)sqrt(y/x)

To proceed it is simpler to look at this as:
f_(x) = 4e^(4x) - (1/2)sqrt(y)*x^(-1/2)

So:
f_(xx) = 16e^(4x) - (1/2)sqrt(y)*(-1/2)x^(-3/2)

Which is the same as:
f_(xx) = 16e^(4x) + (1/4)sqrt(y/x^3)

The other two are done the same way, though they require a bit more work.

-Dan

Hmm, okay, so to do f_(xy), then, we'd take f_(x), which was:

4e^(4x) - (1/2)sqrt(y)*x^(-1/2), and then find the derivative with respect to y, so then the answer is:

0 + {derivative of -(1/2)sqrt(y)*x^(-1/2), treating x^(-1/2) as a constant, here}

So: -1/[4*sqrt(y)*sqrt(x)] ?

And then, to find f_(yyx) can I just find the derivative with respect to y of the answer above and that'll be f_(yyx)?

EDIT: hmm if you do that, you'd get 1/(8*y^(3/2)*sqrt(x)) ...

Originally Posted by Ideasman

Hmm, okay, so to do f_(xy), then, we'd take f_(x), which was:

4e^(4x) - (1/2)sqrt(y)*x^(-1/2), and then find the derivative with respect to y, so then the answer is:

0 + {derivative of -(1/2)sqrt(y)*x^(-1/2), treating x^(-1/2) as a constant, here}

So: -1/[4*sqrt(y)*sqrt(x)] ?

And then, to find f_(yyx) can I just find the derivative with respect to y of the answer above and that'll be f_(yyx)?

This is correct. The f_(yyx) takes a bit of simplification to get it to the form your book has.

-Dan