Worksheet: The Chain Rule for Multivariate Functions

In this worksheet, we will practice using the chain rule with partial derivative of multivariable functions, such as finding ∂ f(x,y) / ∂u, where x=g(u,v) and y=h(u,v).

Q1:

Let 𝑧 = 𝑓 ( 𝑥 , 𝑦 ) over the curve 𝑥 = 𝑡 + 1 , 𝑦 = 𝑡 1 . Write an expression for d d 𝑧 𝑡 , giving your answer in terms of the partial derivatives of 𝑓 .

  • A 2 𝑡 𝜕 𝑓 𝜕 𝑥 ( 𝑡 + 1 , 𝑡 1 ) + 2 𝑡 𝜕 𝑓 𝜕 𝑦 ( 𝑡 + 1 , 𝑡 1 )
  • B 𝑡 𝜕 𝑓 𝜕 𝑥 ( 𝑡 + 1 , 𝑡 1 ) × 𝑡 𝜕 𝑓 𝜕 𝑦 ( 𝑡 + 1 , 𝑡 1 )
  • C 𝑡 𝜕 𝑓 𝜕 𝑥 ( 𝑡 + 1 , 𝑡 1 ) + 𝑡 𝜕 𝑓 𝜕 𝑦 ( 𝑡 + 1 , 𝑡 1 )
  • D 2 𝑡 𝜕 𝑓 𝜕 𝑥 ( 𝑡 + 1 , 𝑡 1 ) × 2 𝑡 𝜕 𝑓 𝜕 𝑦 ( 𝑡 + 1 , 𝑡 1 )
  • E 2 𝑡 𝜕 𝑓 𝜕 𝑥 ( 𝑡 + 1 , 𝑡 1 ) 2 𝑡 𝜕 𝑓 𝜕 𝑦 ( 𝑡 + 1 , 𝑡 1 )

Q2:

The differentiable functions 𝑥 = 𝜙 ( 𝑡 ) , 𝑦 = 𝜓 ( 𝑡 ) describe a curve with ( 𝜙 ( 1 ) , 𝜓 ( 1 ) ) = ( 2 , 3 ) . Use the chain rule to find an expression for d d 𝑤 𝑡 | | | , where 𝑤 = 𝑒 ( 𝑦 ) c o s .

  • A 𝑒 ( 3 ) 𝜙 ( 1 ) + 𝑒 ( 3 ) 𝜓 ( 1 ) c o s s i n
  • B 𝑒 𝜙 ( 1 ) ( 3 ) 𝜓 ( 1 ) s i n
  • C 𝑒 ( 2 ) 𝜙 ( 1 ) 𝑒 ( 2 ) 𝜓 ( 1 ) s i n c o s
  • D 𝑒 ( 3 ) 𝜓 ( 1 ) 𝑒 ( 3 ) 𝜙 ( 1 ) c o s s i n
  • E 𝑒 ( 3 ) 𝜙 ( 1 ) 𝑒 ( 3 ) 𝜓 ( 1 ) c o s s i n

Q3:

Let 𝑓 ( 𝑥 , 𝑦 ) = 𝑥 + 𝑦 and consider the curve given by 𝜙 ( 𝑡 ) = ( 2 𝑡 , 𝑡 ) s i n c o s . Use the chain rule to determine the values of 𝑡 when d d 𝑡 𝑓 ( 𝜙 ( 𝑡 ) ) = 0 . You may keep your answer in terms of c o s .

  • A 𝑛 𝜋 4 , 𝑛 𝜋 1 2 , 𝑛 𝜋 + 1 2 c o s c o s where 𝑛 is an integer
  • B 𝑛 𝜋 4 , 𝑛 𝜋 3 5 , 𝑛 𝜋 + 3 5 c o s c o s where 𝑛 is an integer
  • C 𝑛 𝜋 2 , 𝑛 𝜋 1 4 , 𝑛 𝜋 + 1 4 c o s c o s where 𝑛 is an integer
  • D 𝑛 𝜋 2 , 𝑛 𝜋 1 2 , 𝑛 𝜋 + 1 2 c o s c o s where 𝑛 is an integer
  • E 𝑛 𝜋 2 , 𝑛 𝜋 2 5 , 𝑛 𝜋 + 2 5 c o s c o s where 𝑛 is an integer

Q4:

Let 𝑤 = 𝑓 ( 𝑥 , 𝑦 , 𝑧 ) over the curve 𝑥 = 𝑡 , 𝑦 = 1 𝑡 , 𝑧 = 3 𝑡 . Write an expression for d d 𝑤 𝑡 | | | , giving your answer in terms of the partial derivatives of 𝑓 .

  • A d d 𝑤 𝑡 = 1 4 𝜕 𝑓 𝜕 𝑥 2 , 1 4 , 1 2 + 1 1 6 𝜕 𝑓 𝜕 𝑦 2 , 1 4 , 1 2 + 3 𝜕 𝑓 𝜕 𝑧 2 , 1 4 , 1 2
  • B d d 𝑤 𝑡 = 1 4 𝜕 𝑓 𝜕 𝑥 2 , 1 4 , 1 2 𝜕 𝑓 𝜕 𝑦 2 , 1 4 , 1 2 + 3 𝜕 𝑓 𝜕 𝑧 2 , 1 4 , 1 2
  • C d d 𝑤 𝑡 = 2 𝜕 𝑓 𝜕 𝑥 2 , 1 4 , 1 2 + 1 4 𝜕 𝑓 𝜕 𝑦 2 , 1 4 , 1 2 + 1 2 𝜕 𝑓 𝜕 𝑧 2 , 1 4 , 1 2
  • D d d 𝑤 𝑡 = 1 4 𝜕 𝑓 𝜕 𝑥 2 , 1 4 , 1 2 1 1 6 𝜕 𝑓 𝜕 𝑦 2 , 1 4 , 1 2 + 3 𝜕 𝑓 𝜕 𝑧 2 , 1 4 , 1 2
  • E d d 𝑤 𝑡 = 1 6 3 𝜕 𝑓 𝜕 𝑥 2 , 1 4 , 1 2 + 1 . 3 8 6 𝜕 𝑓 𝜕 𝑦 2 , 1 4 , 1 2 + 2 4 𝜕 𝑓 𝜕 𝑧 2 , 1 4 , 1 2

Q5:

The function 𝐹 ( 𝑥 , 𝑦 ) = 𝑥 + 𝑦 can be considered in terms of the polar coordinates via the transformation 𝑐 ( 𝑟 , 𝜃 ) = ( 𝑥 , 𝑦 ) where 𝑥 = 𝑟 𝜃 c o s and 𝑦 = 𝑟 𝜃 s i n . By considering 𝜕 𝐹 𝜕 𝜃 as the second component of 𝐹 ( 𝑐 ) , or otherwise, write an expression for this partial derivative in terms of 𝑥 and 𝑦 .

  • A 𝜕 𝐹 𝜕 𝜃 = 2 𝑥 𝑦 3 𝑥 𝑦
  • B 𝜕 𝐹 𝜕 𝜃 = 2 𝑥 𝑦 + 3 𝑥 𝑦
  • C 𝜕 𝐹 𝜕 𝜃 = 3 𝑥 𝑦 + 𝑥 𝑦
  • D 𝜕 𝐹 𝜕 𝜃 = 𝑥 𝑦 + 𝑥 𝑦
  • E 𝜕 𝐹 𝜕 𝜃 = 𝑥 𝑦 + 𝑥 𝑦

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.