Question

Find ∂w/∂s and ∂w/∂t using the appropriate Chain Rule.

Answer 1

Answer:

The value of $\frac{\partial w}{\partial s}$ is $e^{3t}-18e^{2s+t}$

The value of $\frac{\partial w}{\partial t}$ is $3e^{3t}-9e^{2s+t}$

The partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial s}$ is $e^{30}-18$

The partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial t}=3(e^{30}-3)$

Step-by-step explanation:

Given that the Function point are $w=y^3-9x^2y$

$x=e^s$, $y=e^t$ and s = -5, t = 10

To find $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial t}$using the appropriate Chain Rule :

$w=y^3-9x^2y$  

Substitute the values of x and y in the above equation we get

$w=(e^t)^3-9(e^s)^2(e^t)$

$w=e^{3t}-9e^{2s}.e^t$

Now  partially differentiating w with respect to s by using chain rule we have

$\frac{\partial w}{\partial ∂s}=e^{3t}-9(e^{2s}).2(e^t)$

$=e^{3t}-18e^{2s}.(e^t)$

$=e^{3t}-18e^{2s+t}$

Therefore the value of $\frac{\partial w}{\partial s}$ is $e^{3t}-18e^{2s+t}$

$w=e^{3t}-9e^{2s}.e^t$

Now  partially differentiating w with respect to t by using chain rule we have

$\frac{\partial w}{\partial t}=e^{3t}.(3)-9e^{2s}(e^t).(1)$

$=3e^{3t}-9e^{2s+t}$

Therefore the value of $\frac{\partial w}{\partial t}$ is $3e^{3t}-9e^{2s+t}$

Now put s-5 and t=10 to evaluate each partial derivative at the given values of s and t :

$\frac{\partial w}{\partial s}=e^{3t}-18e^{2s+t}$

$=e^{3(10}-18e^{2(-5)+10}$

$=e^{30}-18e^{-10+10}$

$=e^{30}-18e^0$

$=e^{30}-18$

Therefore the partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial s}$ is $e^{30}-18$

$\frac{\partial w}{\partial t}=3e^{3t}-9e^{2s+t}$

$=3e^{3(10)}-9e^{2(-5)+10}$

$=3e^{30}-9e{-10+10}$

$=3e^{30}-9e{0}$

$=3e^{30}-9$

$\frac{\partial w}{\partial t}=3(e^{30}-3)$

Therefore the partial derivative at s=-5 and t=10 is $\frac{\partial w}{\partial t}=3(e^{30}-3)$