Question

Use the Chain Rule to find az/as and əz/at.

z =tan(u/v),
u = 4s + 2t,
əz
Əs
дz
at
=
=
v=2s 4t

Answer 1

`(\partial z)/(\partial s)=-(5t)/((s-2t)^2)\sec^2\left((2s+t)/(s-2t)\right)`

`(\partial z)/(\partial t)=(5s)/((s-2t)^2)\sec^2\left((2s+t)/(s-2t)\right)`

Explanation:

Given equations:

`z=\tan\left((u)/(v)\right)`

`u=4s+2t`

`v=2s-4t`

Partial derivatives are the rates at which a multivariable function changes with respect to specific individual variables, while keeping all other variables constant.

To find the partial derivatives δz/δs and δz/δt, first calculate the partial derivatives of u and v with respect to s and t.

To find δu/δs, differentiate the term 4s with respect to s, while treating 2t as a constant:

`(\partial u)/(\partial s) = 4`

To find δu/δt, differentiate the term 2t with respect to t, while treating 4s as a constant:

`(\partial u)/(\partial t) = 2`

To find δv/δs, differentiate the term 2s with respect to s, while treating -4t as a constant:

`(\partial v)/(\partial s) = 2`

To find δv/δt, differentiate the term -4t with respect to t, while treating 2s as a constant:

`(\partial v)/(\partial t) = -4`

Now, calculate the partial derivatives of z with respect to u and v by applying the chain rule:

`\begin{aligned}(\partial z)/(\partial u) &= \sec^2\left((u)/(v)\right)\cdot (\partial )/(\partial u)\left((u)/(v)\right)\\\\&=\sec^2\left((u)/(v)\right)\cdot (1)/(v)\\\\&=(1)/(v)\sec^2\left((u)/(v)\right)\end{aligned}`

`\begin{aligned}(\partial z)/(\partial v) &= \sec^2\left((u)/(v)\right)\cdot (\partial )/(\partial v)\left((u)/(v)\right)\\\\&=\sec^2\left((u)/(v)\right)\cdot -(u)/(v^2)\\\\&=-(u)/(v^2)\sec^2\left((u)/(v)\right)\end{aligned}`

`\hrulefill`

Apply the multivariable chain rule to find δz/δs:

`(\partial z)/(\partial s)=(\partial z)/(\partial u)\cdot (\partial u)/(\partial s)+(\partial z)/(\partial v)\cdot(\partial v)/(\partial s)`

`(\partial z)/(\partial s)=(1)/(v)\sec^2\left((u)/(v)\right)\cdot 4-(u)/(v^2)\sec^2\left((u)/(v)\right)\cdot2`

`(\partial z)/(\partial s)=(4)/(v)\sec^2\left((u)/(v)\right)-(2u)/(v^2)\sec^2\left((u)/(v)\right)`

`(\partial z)/(\partial s)=\sec^2\left((u)/(v)\right)\left[(4)/(v)-(2u)/(v^2)\right]`

`(\partial z)/(\partial s)=\sec^2\left((u)/(v)\right)\left[(4v-2u)/(v^2)\right]`

Plug in the expressions for u and v:

`(\partial z)/(\partial s)=\sec^2\left((4s+2t)/(2s-4t)\right)\left[(4(2s-4t)-2(4s+2t))/((2s-4t)^2)\right]`

`(\partial z)/(\partial s)=\sec^2\left((4s+2t)/(2s-4t)\right)\left[(8s-16t-8s-4t)/((2(s-2t))^2)\right]`

`(\partial z)/(\partial s)=\sec^2\left((4s+2t)/(2s-4t)\right)\left[(-20t)/(4(s-2t)^2)\right]`

`(\partial z)/(\partial s)=\sec^2\left((2s+t)/(s-2t)\right)\left[(-5t)/((s-2t)^2)\right]`

`(\partial z)/(\partial s)=-(5t)/((s-2t)^2)\sec^2\left((2s+t)/(s-2t)\right)`

Therefore, δz/δs is:

`\large\boxed{\boxed{(\partial z)/(\partial s)=-(5t)/((s-2t)^2)\sec^2\left((2s+t)/(s-2t)\right)}}`

`\hrulefill`

Apply the multivariable chain rule to find δz/δt:

`(\partial z)/(\partial t)=(\partial z)/(\partial u)\cdot (\partial u)/(\partial t)+(\partial z)/(\partial v)\cdot(\partial v)/(\partial t)`

`(\partial z)/(\partial t)=(1)/(v)\sec^2\left((u)/(v)\right)\cdot 2-(u)/(v^2)\sec^2\left((u)/(v)\right)\cdot(-4)`

`(\partial z)/(\partial t)=(2)/(v)\sec^2\left((u)/(v)\right)+(4u)/(v^2)\sec^2\left((u)/(v)\right)`

`(\partial z)/(\partial t)=sec^2\left((u)/(v)\right)\left[(2)/(v)+(4u)/(v^2)\right]`

`(\partial z)/(\partial t)=sec^2\left((u)/(v)\right)\left[(2v+4u)/(v^2)\right]`

Plug in the expressions for u and v:

`(\partial z)/(\partial t)=\sec^2\left((4s+2t)/(2s-4t)\right)\left[(2(2s-4t)+4(4s+2t))/((2s-4t)^2)\right]`

`(\partial z)/(\partial t)=\sec^2\left((4s+2t)/(2s-4t)\right)\left[(4s-8t+16s+8t)/((2(s-2t))^2)\right]`

`(\partial z)/(\partial t)=\sec^2\left((4s+2t)/(2s-4t)\right)\left[(20s)/(4(s-2t)^2)\right]`

`(\partial z)/(\partial t)=\sec^2\left((2s+t)/(s-2t)\right)\left[(5s)/((s-2t)^2)\right]`

`(\partial z)/(\partial t)=(5s)/((s-2t)^2)\sec^2\left((2s+t)/(s-2t)\right)`

Therefore, δz/δt is:

`\large\boxed{\boxed{(\partial z)/(\partial t)=(5s)/((s-2t)^2)\sec^2\left((2s+t)/(s-2t)\right)}}`