How do you differentiate #x^3 yz^2+w^2 x^2 y-x^3+wsiny=44#?
1 Answer
Explanation:
I'll assume that
You can calculate the partial derivative of
When you differentiate with respect to
On the other hand,
So, let's start with the partial derivative of
Now, you can calculate
#(del)/(delx)(yx^2w^2) = y * [d/dx(x^2) * w^2 + x^2 * (del)/(delx)(w^2)]#
#(del)/(delx)(yx^2w^2) = y * (2x * w^2 + x^2 * 2w * (delw)/(delx))#
#(del)/(delx)(yx^2w^2) = 2xyw^2 + 2x^2yw(delw)/(delx)#
Plug this back into the calculation with respect to
#yz^2 * 3x^2 + 2xyw^2 + 2x^2yw(delw)/(delx) - 3x^2 + siny * (delw)/(delx) = 0#
Rearrange this to isolate
#(delw)/(delx)(2x^2yw + siny) = +3x^2 - 2xyw^2 - 3x^2yz^2#
#(delw)/(delx) = color(green)((3x^2(1 - yz^2) - 2xyw^2)/(2x^2yw + siny))#
Next, take the partial derivative of
#(del)/(dely)(x^3yz^2 + w^2x^2y - x^3 + wsiny) = d/(dy)(44)#
This will get you
#x^3z^2 + x^2 * [(del)/(dely)(w^2) * y + w^2 * d/(dy)(y)] - 0 + [(del)/(dely)(w) * siny + w * d/(dy)(siny)] = 0#
#x^3z^2 + x^2 * (2w * (delw)/(dely) + w^2) + (delw)/(dely) * siny + w * cosy = 0#
#x^3z^2 + 2x^2w(delw)/(dely) + x^2w^2 + (delw)/(dely) * siny + wcosy = 0#
Once again, rearrange to get
#(delw)/(dely)(2x^2w + siny) = -x^3z^2 - x^2w^2 - wcosy#
#(delw)/(dely) = color(green)(-(x^2(xz^2 + w^2) + wcosy)/(2x^2 + siny))#
Finally, take the partial derivative of
#(del)/(delz)(x^3yz^2 + w^2x^2y - x^3 + wsiny) = d/(dz)(44)#
#2x^3yz + 2x^2yw(delw)/(delz) - 0 + siny * (delw)/(delz) = 0#
Rearrange to get
#(delw)/(delz)(2x^2yw + siny) = -2x^3yz#
#(delw)/(delz) = color(green)(-(2x^3yz)/(2x^2yw + siny))#