# Math Help - Partial derivative question

1. ## Partial derivative question

(a) If
f(x, y) = log(x + y) + sin(xy) + y^2e^x, determine partial df/dx and partial df/dt

find df/dt in terms of
x and y where x = cos t, y = sin t using the chain rule formula. [ 10 marks ]

(b) If
f(x, y) = x3 y2 xy, determine partial df/dx and p-artial df/dy

If f = 0 then express dy/dx in terms of x and y.

Just a quick answer check for a i got:
df/dt=(1/x+y+ycos(xy)+y^2e^x)*-y+(1/x+y+ycos(xy)+y^2e^x)*x
Now im sure this isnt in the simpliest form

For part b i got dy/dx= (-3x^2+y)/(-2y-x)?

2. In Question (a) are you sure you're not meant to find $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$, not $\displaystyle \frac{\partial f}{\partial t}$?

3. Sorry yes thats correct need to find partial df/dx and partial df/dy then from that df/dt.

4. Use standard derivative rules, treating all variables as constants except for the variable you are differentiating with respect to...

5. Originally Posted by Prove It
Use standard derivative rules, treating all variables as constants except for the variable you are differentiating with respect to...
Yes i done that to obtain partial df/dx= 1/(x+y)+cos(xy)*y+y^2*e^x then for df/dy= 1/(x+y)+cosy(xy)*x+2y*e^x
Then am i correct to think you can work out dx/dt and dx/dy then to multiply the partial df/dx by dx/dt and partial df/dy by dy/dt to form df/dt?

6. Yes, your partial derivatives are correct.

Now use the chain rule: $\displaystyle \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\,\frac{dx}{dt} + \frac{\partial f}{\partial y}\,\frac{dy}{dt}$.

Alternatively, you could substitute $\displaystyle x = \cos{t}$ and $\displaystyle y = \sin{t}$ into $\displaystyle f(x,y)$ to get a function $\displaystyle f(t)$.

7. Originally Posted by Prove It
Yes, your partial derivatives are correct.

Now use the chain rule: $\displaystyle \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\,\frac{dx}{dt} + \frac{\partial f}{\partial y}\,\frac{dy}{dt}$.

Alternatively, you could substitute $\displaystyle x = \cos{t}$ and $\displaystyle y = \sin{t}$ into $\displaystyle f(x,y)$ to get a function $\displaystyle f(t)$.
I think thats what i may of done with dx/dt=-sint=-y and dy/dt=cost=x therefore getting: (1/x+y+ycos(xy)+y^2e^x)*-y+(1/x+y+ycos(xy)+y^2e^x)*x

8. Originally Posted by breitling
I think thats what i may of done with dx/dt=-sint=-y and dy/dt=cost=x therefore getting: (1/x+y+ycos(xy)+y^2e^x)*-y+(1/x+y+ycos(xy)+y^2e^x)*x
Looks good to me