Revision history [back]

The way the problem is state, I believe you simply swap out $x$ and $y$ for their $x(s,t)$ and $y(x,t)$ counterparts thus generating the equation:

$$ z = x^2y $$

Substituting here for $x$ and $y$:

$$ z = sin^2(st)(t^2 + s^2) $$

Now it should be a simple matter of taking the partial derivatives (with quite a bit of chain rule in here)

$$ \frac {\partial z} {\partial t} = 2tsin^2(st) + (t^2 + s^2)(2s*sin(st)cos(st)) $$

And then with respect to s

$$ \frac {\partial z} {\partial s} = 2ssin^2(st) + (t^2 + s^2)(2t*sin(st)cos(st)) $$

The trick to this question is recognizing that $x$ and $y$ are their own functions of 's' and 't' and plugging this into the equation given to create the new function $z(s, t)$.

Hope this helps

The way the problem is state, I believe you simply swap out $x$ and $y$ for their $x(s,t)$ and $y(x,t)$ counterparts thus generating the equation:

$$ z = x^2y $$

Substituting here for $x$ and $y$:

$$ z = sin^2(st)(t^2 + s^2) $$

Now it should be a simple matter of taking the partial derivatives (with quite a bit of chain rule in here)

$$ \frac {\partial z} {\partial t} = 2tsin^2(st) + (t^2 + s^2)(2s*sin(st)cos(st)) $$

And then with respect to s

$$ \frac {\partial z} {\partial s} = 2ssin^2(st) + (t^2 + s^2)(2t*sin(st)cos(st)) $$

The trick to this question is recognizing that $x$ and $y$ are their own functions of 's' and 't' and plugging this into the equation given to create the new function $z(s, t)$.

Hope this helps

currently editing-Comparing these answers back to the ones in the book, instead of $2s2x$ i have calculated $2sx^2$. I am not sure if I am wrong or if the book is wrong in this instance, but i calculated the partial derivative on a calculator and i got the answer i have provided-

The way the problem is state, I believe you simply swap out $x$ and $y$ for their $x(s,t)$ and $y(x,t)$ counterparts thus generating the equation:

$$ z = x^2y $$

Substituting here for $x$ and $y$:

$$ z = sin^2(st)(t^2 + s^2) $$

Now it should be a simple matter of taking the partial derivatives (with quite a bit of chain rule in here)

$$ \frac {\partial z} {\partial t} = 2tsin^2(st) + (t^2 + s^2)(2s*sin(st)cos(st)) $$

And then with respect to s

$$ \frac {\partial z} {\partial s} = 2ssin^2(st) + (t^2 + s^2)(2t*sin(st)cos(st)) $$

When we simplify this back to $x$ and $y$ terms we get the equations:

$$ \frac {\partial z} {\partial s} = 2sx^2 + 2txtcos(st) $$

and

$$ \frac {\partial z} {\partial t} = 2tx^2 + 2sxycos(st) $$

The trick to this question is recognizing that $x$ and $y$ are their own functions of 's' and 't' and plugging this into the equation given to create the new function $z(s, t)$.

Hope this helps

-Comparing these answers back to the ones in the book, instead of $2s2x$ i have calculated $2sx^2$. I am not sure if I am wrong or if the book is wrong in this instance, but i calculated the partial derivative on a calculator and i got the answer i have provided-

The way the problem is state, I believe you simply swap out $x$ and $y$ for their $x(s,t)$ and $y(x,t)$ counterparts thus generating the equation:

$$ z = x^2y $$

Substituting here for $x$ and $y$:

$$ z = sin^2(st)(t^2 + s^2) $$

Now it should be a simple matter of taking the partial derivatives (with quite a bit of chain rule in here)

$$ \frac {\partial z} {\partial t} = 2tsin^2(st) + (t^2 + s^2)(2s*sin(st)cos(st)) $$

And then with respect to s

$$ \frac {\partial z} {\partial s} = 2ssin^2(st) + (t^2 + s^2)(2t*sin(st)cos(st)) $$

When we simplify this back to $x$ and $y$ terms we get the equations:

$$ \frac {\partial z} {\partial s} = 2sx^2 + 2txtcos(st) $$

and

$$ \frac {\partial z} {\partial t} = 2tx^2 + 2sxycos(st) $$

The trick to this question is recognizing that $x$ and $y$ are their own functions of 's' and 't' and plugging this into the equation given to create the new function $z(s, t)$.

Hope this helps

-Comparing these answers back to the ones in the book, instead of $2s2x$ i have calculated $2sx^2$. I am not sure if I am wrong or if the book is wrong in this instance, but i calculated the partial derivative on a calculator and i got the answer i have provided-

The way the problem is state, I believe you simply swap out $x$ and $y$ for their $x(s,t)$ and $y(x,t)$ counterparts thus generating the equation:

$$ z = x^2y $$

Substituting here for $x$ and $y$:

$$ z = sin^2(st)(t^2 + s^2) $$

Now it should be a simple matter of taking the partial derivatives (with quite a bit of chain rule in here)

$$ \frac {\partial z} {\partial t} = 2tsin^2(st) 2tsin^2(st) + (t^2 + s^2)(2s*sin(st)cos(st)) $$ s^2)(2ssin(st)cost(st)) $$

And then with respect to s

s $$ \frac {\partial z} {\partial s} = 2ssin^2(st) 2ssin^2(st) + (t^2 + s^2)(2t*sin(st)cos(st)) s^2)(2tsin(st)cos(st)) $$

When we simplify this back to $x$ and $y$ terms we get the equations:

$$ \frac {\partial z} {\partial s} = 2sx^2 + 2txtcos(st) 2sx^2 + 2txycos(st) $$

and

$$ \frac {\partial z} {\partial t} = 2tx^2 + 2sxycos(st) 2tx^2 + 2sxycos(st) $$

The trick to this question is recognizing that $x$ and $y$ are their own functions of 's' and 't' and plugging this into the equation given to create the new function $z(s, t)$.

Hope this helps