# Let z = x^2 y^3 , where x = 4s + 3t and y = 5s + 2t. Find \frac{\partial z}{\partial s} and ...

## Question:

Let {eq}z = x^2 y^3 {/eq}, where x = 4s + 3t and y = 5s + 2t. Find {eq}\frac{\partial z}{\partial s} {/eq} and {eq}\frac{\partial z}{\partial t} {/eq} for s = -1, t = 2. (The answer should consist of two number.)

## Partial Derivative:-

If the question is asking for finding partial derivative and the giving the values of independent variable.

Then firstly find the derivative of the function using the chain rule and then find its value by putting the values of the given values of independent variable. After that we got the answer as a numeric value.

Usual Formula:-

{eq}\boxed{\dfrac{\partial z}{\partial s}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial s}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial s}}{/eq}

Given:

{eq}z = x^{2}y^{3}\\ x = 4s+3t\\ y=5s+2t {/eq}

then {eq}\dfrac{\partial x}{\partial s}=4 ,\dfrac{\partial x}{\partial t}=3\\ \dfrac{\partial y}{\partial s}=5 , \dfrac{\partial y}{\partial t}=2 {/eq}

And also

{eq}\dfrac{\partial z}{\partial x}=2xy^{3} ,\dfrac{\partial z}{\partial y}=3x^{2}y^{2} {/eq}

Now {eq}\dfrac{\partial z}{\partial s}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial s}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial s}\\ \dfrac{\partial z}{\partial s}=2xy^{3}(4)+3x^{2}y^{2}(5)\\ \dfrac{\partial z}{\partial s}=8(4s+3t)(5s+2t)^{3}+15(4s+3t)^{2}(5s+2t)^{2} {/eq}

At s =-1 and t =2:-

{eq}\dfrac{\partial z}{\partial s}=8(4(-1)+3(2))(5(-1)+2(2))^{3}+15(4(-1)+3(2))^{2}(5(-1)+2(2))^{2}\\ \dfrac{\partial z}{\partial s}=8(2)(-1)+15(4)(1)\\ \dfrac{\partial z}{\partial s}=-16+60=44 {/eq}

And {eq}\dfrac{\partial z}{\partial t}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial t}\\ \dfrac{\partial z}{\partial t}=2xy^{3}(3)+3x^{2}y^{2}(2)\\ \dfrac{\partial z}{\partial t}=6(4s+3t)(5s+2t)^{3}+6(4s+3t)^{2}(5s+2t)^{2} {/eq}

At s =-1 and t =2:-

{eq}\dfrac{\partial z}{\partial t}=6(4(-1)+3(2))(5(-1)+2(2))^{3}+6(4(-1)+3(2))^{2}(5(-1)+2(2))^{2}\\ \dfrac{\partial z}{\partial t}=6(2)(-1)+6(4)(1)\\ \dfrac{\partial z}{\partial t}=-12+24=12 {/eq}