Question

Use the chain rule to find the indicated partial derivatives. Z = x4 + x2y, x = s + 2t − u, y = stu2; ∂z ∂s , ∂z ∂t , ∂z ∂u when s = 4, t = 5, u = 1

Answer 1

`z= x^(4)+ x^(2)y\\x= s+2t-u\\y= stu^(2)`

to proof - by using the chain rule.

`(\partial z)/(\partial s) = (\partial z)/(\partial x) * (\partial x)/(\partial s) + (\partial z)/(\partial y) * (\partial y)/(\partial s)`

=
`= \left ( 4x^(3) +2xy \right ).1 + x^(2).\left ( tu^(2) \right )`

put the value of z, x and y.

we get

`= \left ( s +2t-u \right )\left ( tu^(2)\left ( s+2t-u \right )+2stu^(2)+4\left ( s+2t-u \right )^(2) \right )`

now put the value of s=4, t= 5, u=1 in the above question.

now by putting the value we get

= 14236

now we find the

`(\partial z)/(\partial t) = (\partial z)/(\partial x) * (\partial x)/(\partial t) + (\partial z)/(\partial y) * (\partial y)/(\partial t)`

after doing differentiation

we get

`\left ( 4x^(3)+2xy \right )* 2 + x^(2)* \left ( su^(2) \right )`

now the put the value of x,y,z

we get

`=su^(2)\left ( s+2t-u \right )^(2) + 2 \left ( 2stu^(2)\left ( s + 2t - u \right )+ 4\left ( s + 2t-4 \right )^(3) \right )`

now we put the value of s=4, t=5,u=1

=19292

hence proved

now we find the value

`(\partial z)/(\partial u) = (\partial z)/(\partial x) * (\partial x)/(\partial u) + (\partial z)/(\partial y) * (\partial y)/(\partial u)`

`=\left ( 4x^(3)+2xy \right )\left ( -1 \right ) +\left ( x^(2) \right ).\left ( 2stu \right )`

now put the value of x, y and z.

`=2\left ( s+2t-u \right )(-stu^(2)-2\left ( s+2t-u \right )^(2) + stu (s+2t-u))`

now put the value of u, v, z

we get

= -2548

hence proved