# Let W(s, t) = F(u(s, t), v(s, t)), where F, u, and v are differentiable, and the following applies.

–  $u( \space – \space 9, \space 6 ) \space = \space – \space 6 , \space v ( \space – 9, \space 6 ) = \space – \space 4$.

–  $u_s( \space – \space 9, \space 6 ) \space = \space – \space 6 , \space v_t ( \space – 9, \space 6 ) = \space 5$.

–  $u_t( \space – \space 9, \space 6 ) \space = \space – \space 6 , \space v_t( \space – 9, \space 6 ) = \space – \space 5$.

–  $F_u( \space – \space 9, \space 6 ) \space = \space – \space 6 , \space F_v ( \space – 9, \space 6 ) = \space 4$.

Find $W_s(- space 9, \space 6 )$ and  $W_t(- space 9, \space 6 )$.

The main objective of this question is to find the value of given function using chain rule.

This question uses the concept of chain rule to find the value of given function. The chain rule explains how the derivative of the sum of two differentiable functions can be written in terms of the derivatives of those two functions.

We know that:

$\space \frac{ dW }{ ds } \space = \space \frac{ dW }{ du } \space . \space \frac{ du }{ ds } \space +\space \frac{ dW }{ dv } \space . \space \frac{ dv }{ ds }$

By substituting the values, we get:

$\space W_s(- space 9, \space 6) \space = \space F_u( – space 6, \space – \space 4 ) \space . \space u_s( – space 9, \space 6 ) \space + \space F_v( – space 6, \space 4 ) \space . \space v_S( – space 6, \space 4 )$

$\space = \space 0 \space + \space 20$

$\space = \space 20$

Hence, $W_s(- \space 9, \space 6)$ is $20$.

Now using the chain rule for $W_t(s,t)$, so:

$\space \frac{ dW }{ dt } \space = \space \frac{ d}{ dW } \space . \space \frac{ du }{ dt } \space +\space \frac{ dW }{ dv } \space . \space \frac{ dv }{ dt }$

By substituting the values, we get:

$\space W_t(- space 9, \space 6) \space = \space F_u( – space 6, \space – \space 4 ) \space . \space u_t( – space 9, \space 6 ) \space + \space F_v( – space 6, \space 4 ) \space . \space v_t( – space 6, \space 4 )$

$\space =\space 16 \space – \space 20$

$\space = \space – \space 6$

Hence, $W_t(- \space 9, \space 6)$ is $- 6$.

The value of $W_s(- \space 9, \space 6)$ is $20$.

The value of $W_t(- \space 9, \space 6)$ is $- 6$.

## Example

In the above question, if:

• $\space u(1, −9) =3$
• $\space v(1, −9) = 0$
• $\space u_s(1, −9) = 9$
• $\space v_s(1, −9) = −6$
• $\space u_t(1, −9) = 4$
• $\space v_t(1, −9) = 7$
• $\space F_u(3, 0) = −2$
• $\space F_ v(3, 0) = −4$

Find W_s(1, −9) and W_t(1, −9).

For finding  $W_s$, we have:

$\space W(s, t) \space = \space F(u(s, t), v(s, t))$

$\space (1,-9) \space = \space((u(1, -9), v(1, -9)), (u(1, -9), v(1, -9))) · ((1, -9), (1, -9))$

By substituting the values, we get:

$\space = \space 6$

Now for finding $W_t$, we have:

$\space = \space (F_u (3, 0), F_v (3, 0)) · (4, 7)$

$\space = \space – \space 36$