Partial Derivatives Practice

Questions

  1.  

∂∂x(xeyz)= ?\frac{\partial}{\partial x}\big(xe^{yz}\big)=\ ?

  1.  

∂∂y(xeyz)= ?\frac{\partial}{\partial y}\big(xe^{yz}\big)=\ ?

  1.  

∂∂x ln⁡(x2+y2)= ?\frac{\partial}{\partial x}\,\ln(x^2+y^2)=\ ?

  1.  

∂∂y (x2y3+sin⁡(xy))= ?\frac{\partial}{\partial y}\,\big(x^2y^3+\sin(xy)\big)=\ ?

  1.  

∂2∂x ∂y exy= ?\frac{\partial^2}{\partial x\,\partial y}\,e^{xy}=\ ?

  1.  

∂∂x (x+yz)3= ?\frac{\partial}{\partial x}\,(x+yz)^{3}=\ ?

  1.  

∂∂x (y2cos⁡(xy))= ?\frac{\partial}{\partial x}\,\big(y^{2}\cos(xy)\big)=\ ?

  1.  

∂∂x xyzx+y+z= ?\frac{\partial}{\partial x}\,\frac{xyz}{x+y+z}=\ ?

  1. Evaluate at the point

f(x,y)=x2+y2,∂f∂y(3,4)= ?f(x,y)=\sqrt{x^{2}+y^{2}},\qquad \frac{\partial f}{\partial y}(3,4)=\ ?

 

Solutions

1)

Partial derivatives are not as hard as they look!

Basically, just differentiate with respect to the given variable and treat the other variables as constants!

For example,

∂∂x(x2+8y)=2x\frac{\partial}{\partial x} (x^2 + 8y) = 2x ∂∂y(x2+8y)=8\frac{\partial}{\partial y} (x^2 + 8y) = 8

Therefore,

∂∂x(xeyz)=eyz\frac{\partial}{\partial x} (xe^{yz}) = e^{yz}

 

2)
Treat x,zx,z as constants; chain rule on eyze^{yz}:

∂∂y(xeyz)=x⋅eyz⋅z=xz eyz.\frac{\partial}{\partial y}\big(xe^{yz}\big) = x\cdot e^{yz}\cdot z = xz\,e^{yz}.

 

3)

∂∂xln⁡(x2+y2)=1x2+y2⋅2x=2xx2+y2.\frac{\partial}{\partial x}\ln(x^2+y^2) =\frac{1}{x^2+y^2}\cdot 2x =\frac{2x}{x^2+y^2}.

 

4)

∂∂y(x2y3+sin⁡(xy))=x2⋅3y2+cos⁡(xy)⋅x=3x2y2+xcos⁡(xy).\frac{\partial}{\partial y}\big(x^2y^3+\sin(xy)\big) = x^2\cdot 3y^2 + \cos(xy)\cdot x = 3x^2y^2 + x\cos(xy).

 

5)
First ∂/∂y \partial/\partial y: ∂∂yexy=xexy\frac{\partial}{\partial y}e^{xy}=x e^{xy}.
Then ∂/∂x \partial/\partial x:

∂∂x(xexy)=exy+xy exy=exy(1+xy).\frac{\partial}{\partial x}\big(x e^{xy}\big) = e^{xy}+xy\,e^{xy} = e^{xy}(1+xy).

 

6)
Hold y,zy,z fixed:

∂∂x(x+yz)3=3(x+yz)2.\frac{\partial}{\partial x}(x+yz)^3 =3(x+yz)^2.

 

7)
Product/chain with yy treated constant:

∂∂x(y2cos⁡(xy))=y2(−sin⁡(xy))⋅y=−y3sin⁡(xy).\frac{\partial}{\partial x}\big(y^2\cos(xy)\big) = y^2\big(-\sin(xy)\big)\cdot y = -y^{3}\sin(xy).

 

8)
Quotient; N=xyz, D=x+y+zN=xyz,\ D=x+y+z.
Nx=yz, Dx=1N_x=yz,\ D_x=1.

∂∂xxyzx+y+z=(yz)(x+y+z)−xyz⋅1(x+y+z)2=yz(y+z)(x+y+z)2.\frac{\partial}{\partial x}\frac{xyz}{x+y+z} =\frac{(yz)(x+y+z)-xyz\cdot 1}{(x+y+z)^2} =\frac{yz(y+z)}{(x+y+z)^2}.

 

9)

fy=yx2+y2  ⇒  fy(3,4)=432+42=45.f_y=\frac{y}{\sqrt{x^2+y^2}}\;\Rightarrow\; f_y(3,4)=\frac{4}{\sqrt{3^2+4^2}}=\frac{4}{5}.

 

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