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.

$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \times \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \times \frac{\partial y}{\partial s}$

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

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

$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \times \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \times \frac{\partial y}{\partial t}$

after doing differentiation

we get

$\left ( 4x^{3}+2xy \right )\times 2 + x^{2}\times \left ( su^{2} \right )$

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

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \times \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \times \frac{\partial y}{\partial u}$

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

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